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THE  ETHEL  CARR  PEACOCK 

MEMORIAL  COLLECTION 


Matris  amori  monumentum 


TRINITY  COLLEGE  LIBRARY 


DURHAM,  N.  C. 

1903 

Gift  of  Dr.  and  Mrs.  Dred  Peacock 


Digitized  by  the  Internet  Archive 
in  2016  with  funding  from 
Duke  University  Libraries 


https://archive.org/details/elementarypracti01dodd 


L 


ELEMENTARY  AND  PRACTICAL 

ALGEBRA: 

IN  WHICH  HAVE  BEEN  ATTEMPTED 

IMPROVEMENTS 

IN  GENERAL  ARRANGEMENT  AND  EXPOSITION: 


AND  IN  THE  MEANS  OF  THOROUGH  DISCIPLINE  IN  THE  PRINCIPLES  AND 
APPLICATIONS  OF  THE  SCIENCE. 


BY  JAMES  B.  DODD,  A.  M., 

Morrison  Professor  of  Mathematics  and  Natural  Philosophy  in  Transylvania 

University. 


FOURTEENTH  EDITION. 

NEW  YORK: 

PRATT,  OAKLEY  & COMPANY, 


21  MURRAY  STREET. 


ISfiO. 


Entered,  according  to  Act  of  Congress,  in  tne  year-  1852, 
By  JAMES  B.  DODD, 

In  the  Clerk’s  Office  of  the  District  Court  of  Kentucky. 


I.  P.  JONES  i.  CO.,  STEREOTYPEaS. 
I S3  William  street 


S'12,.02. 

0U5°, 

P I 

PREFACE. 


The  following  work  is  designed  to  furnish,  a practicable 
course  of  Algebra  for  the  younger  classes  of  students,  without 
the  omission  of  any  thing  important  to  a thorough  education 
in  the  subjects  which  it  embraces. 

It  aims  at  the  most  methodical  arrangement , the  clearest 
expositions , the  best  elementary  exercises , and  the  most  varied 
and  useful  applications  : — in  all  these  respects  presenting  some 
new  features , which  have  been  adopted  as  improvements  in  the 
method  of  teaching  this  science. 


All  that  is  appropriate  to  an  Algebraic  treatise  in  a general 
course  of  mathematical  studies,  or  necessary  in  preparation  for 
the  higher  Avorks  of  the  course,  has  been  introduced,  with  the 
exception  of  a few  subjects  Avhich  are  more  exclusively  pre- 
liminary to  the  Differential  and  Integral  Calculus.  These, 
with  whatever  else  may  be  considered  useful  in  a larger  work, 
will  shortly  be  added  to  the  present  treatise. 

Between  this  work  and  the  author’s  Arithmetic  there  will 
be  found  a mutual  correspondence  in  many  respects,  though 
each  is  entirely  complete  in  itself.  The  two  are  commended 
% to  the  consideration  of  Teachers  of  Mathematics,  and  the 
Guardians  and  Friends  of  Education,  as  containing  a practi- 
cable, progressive,  and  thorough  course  of  study,  for  Schools, 
on  these  connected  and  important  branches  of  science. 


Transylvania  University, 
July  20th,  1852. 


234^7 


REMARKS 


ON  THE  METHOD  OF  USING  THIS  WORK,  AND  CONDUCTING 
EXAMINATIONS  IN  ALGEBRA. 


The  following  remarks  may  be  useful  to  the  less  experienced  Teacher  using  this 
work,  who  would  make  it  fully  efficient  for  the  purposes  intended. 

1.  The  definitions  and  propositions  numbered  (1),  (2),  (3),  <£c,  and  the  Rules  L 
II,  III,  <Sec,  should  be  accurately  memorized  and  recited  by  the  Student 

2.  The  accompanying  examples,  illustrations,  or  demonstrations,  should  be  required 
of  the  Student,  and  discussed  with  him  on  the  part  of  the  Teacher,  with  reference  to 
the  principles  involved  in  them. 

3.  The  oral  exercises  in  the  earlier  parts  of  the  work,  should  be  exacted;  and  the 
Student  should  often  be  examined  on  the  exercises  under  the  Rules,  with  his  book 
closed. 

4.  In  the  solution  of  Equations  and  Problems,  he  should  explain  each  part  of  the 
operation,  as  exemplified  in  different  pails  of  the  work. 

5.  The  Analysis  of  Contents  (see  the  next  page)  will  be  convenient  for  reviews 
on  the  theory  of  the  science ; and  such  reviews  should  be  frequent.  The  Student  will 
thus  become  familiar  with  the  phraseology,  principles,  and  order  of  the  science. 

6.  The  Student’s  acquisitions  will  depend  very  much  on  the  exactness,  as  well  as 
on  the  frequency,  with  which  he  is  examined.  The  requisitions  made  on  him  should 
be  adapted  to  his  capabilities, — which,  it  should  be  remembered,  are  liable  to  be 
sometimes  overrated,  and  sometimes  underrated,  by  Authors  and  Teachers. 


ANALYSIS  OF  CONTENTS 


5^p’=0  This  Analysis  is  designed  to  be  used  in  oral  examinations,  in  reviews.  The 
Teacher  will  name  the  topic  as  presented  in  this  table ; the  Learner  will  respond 
according  to  his  knowledge  of  the  subject. 

Tor  example:  the  Teacher  will  say,  “Science  and  Art;”  the  Learner  will  re- 
spond, “ Science  is  knowledge  reduced  to  a system  ; Art  is  knowledge  applied  to 
practical  purposes.” 


CHAPTER  I. 

Preliminary  Definitions  and  Exercises. — Page  1 ...  8. 

Science  and  Art,  (1). — A Unit — Numbers.  (2). — Quantity — Whether 
Numbers  are  quantities , (3). — Mathematics — Its  most  general  Divisions — 
Arithmetic — Geometry,  (4). — Algebra,  (5). — Symbols  of  Quantities,  (6). — 
Symbols  of  Operations,  the  Sign  -|-  plus,  (7). — The  Sign  — minus,  (8). — The 
Sign  x into — A point  (. ) between  Quantities — Quantities  in  juxtaposition, 

(9) . — The  sign  y bp — Division  otherwise  denoted — An  Integral  Quantity, 

(10) . — Use  of  the  Parenthesis  or  Vinculum,  (11). — Factors  and  Constant 
Produet,  (12). — Powers  of  Quantities,  (13). — Roots  of  Quantities,  (14). — 
The  Coefficient  of  a Quantity,  (15). — The  Exponent  of  a Quantity,  (16). — 
What  an  Integral  Coefficient  indicates — an  Integral  Exponent,  (17). — Simi- 
lar and  Dissimilar  Quantities,  (18). — An  Algebraic  Monomial,  (19). — An 
Algebraic  Polynomial — A Binomial — A Trinomial,  (20). — Whether  the 
Value  of  a Polynomial  is  affected  by  changing  the  Order  of  its  Terms.  (21). 
— A Polynomial  arranged  by  powers,  (22). — A Homogeneous  Polynomial — 
Dimensions  and  Degrees  of  the  Terms  of  a Polynomial,  (23). — Positive  and 
Negative  Quantities,  (24). — Effect  of  a Negative  Quantity  in  an  Expression 
or  a Calculation — Occasional  Use  of  the  Positive  and  Negative  Signs.  (25) 


VI 


ANALYSIS  OF  CONTENTS. 


CHAPTER  II. 

Addition. — Subtraction. — Multiplication. — Division. — 9 ...  26. 

Algebraic  Addition,  (26). — How  to  Add  Similar  Terms  with  like  signs, 
(27). — Sum  of  Two  Equal  Similar  Terms  with  contrary  signs,  (28). — How 
to  Add  unequal  Similar  Terms  with  contrary  signs,  (29)  — Sum  of  Two  oi 
more  Dissimilar  Terms,  (30). 

Rule  I.  For  the  Addition  of  Algebraic  Quantities , (31). 

Calculations  on  the  same  Polynomial  in  two  or  more  ( )’s,  (32). 

Algebraic  Subtraction,  (33). — How  to  Subtract  a Monomial  from 
another  Quantity,  (34). — How  to  Subtract  a Monomial  from  a Dissimilar 
Quantity,  (35). 

Rule  II.  For  the  Subtraction  of  Algebraic  Quantities , (36). 

How  to  denote  the  Subtraction  of  a negaiive  Monomial — of  a Polyno- 
mial, (37). — Change  of  signs  in  a Polynomial  without  affecting  its  Value, 
(38). 

Algebraic  Multiplication — When  the  Multiplier  is  positive — When  the 
Multiplier  is  negative , (39). — Product  of  Two  Monomials,  (40). — Exponent, 
in  the  Product,  of  a Letter  occurring  in  both  the  Monomials  multiplied  to- 
gether, (41). — Sign  of  the  Product — Reason  for  this  when  both  the  Quanti- 
ties are  negative — When  One  of  them  is  positive  and  the  Other  negative 
(42). — Product  when  Either  of  the  Two  Factors  is  0,  (43). 

Rule  III.  To  Multiply  a Monomial  into  a Polynomial,  (44). 

Rule  IV.  To  Multiply  a Polynomial  into  a Polynomial . (45). 

Algebraic  Division,  (46). — Howto  find  the  Quotient  of  Two  Monomials, 
(47). — Value  of  any  Quantity  with  exponent  0.  (48). — Sign  of  the  Quotient, 
and  Principle  which  determines  it,  (49) — Quotient  of  0 divided  by  any 
Quantity,  and  of  any  Quantity  divided  by  0,  (50). 

Rule  V.  To  Divide  a Monomial  into  a Polynomial,  (51). 

Rule  VI.  To  Divide  a Polynomial  into  a Polynomial,  (52). 

How  the  indicated  Product  of  Two  or  more  Factors  may  be  divided.  (5oy 


ANALYSIS  OF  CONTENTS. 


•vii 

CHAPTER  III. 

Composite  Quantities. — Common  Measure. — Common  Multiple. — 

27  . . . 40. 

A Composite  and  a Prime  Quantity,  (54). — Decomposition  of  a Quantity 
— Divisor  and  Quotient  as  Factors  of  a Quantity,  (55). — What  the  Differ- 
ence of  Two  Quantities  will  divide,  (56). — What  the  sum  of  two  Quantities 
will  divide,  (57). — Product  of  the  Sum  and  Difference  of  two  Quantities, 
(58). — Square  of  the  Sum  of  two  Quantities,  (59). — Square  of  the  Difference 
of  two  Quantities,  (60). — Case  in  which  a Trinomial  may  be  resolved  into 
Two  unequal  Binomial  Factors,  (61.) — One  Quantity  a Measure  of  another 
— A Common  Measure  of  Two  or  more  Quantities,  (62). — Greatest  Common 
Measure  of  Two  or  More  Quantities,  (63). — Composition  of  the  Greatest 
Common  Measure,  (64). — Principle  on  which  depends  the  Rule  for  the 
Greatest  Common  Measure,  (65). 

Rule  VII. — To  Find  the  Greatest  Common  Measure  of  Two  Quantities,  (66). 

Whether  the  Signs  in  a Common  Measure  may  be  changed,  (67). — Of  a 
Factor  which  is  common  to  all  the  Terms  of  the  Dividend,  (68). — Of  a 
Factor  which  is  contained  in  all  the  Terms  of  the  two  Polynomials,  (69). — 
One  Quantity  a Multiple  of  another. — A Common  Multiple  of  Two  or 
more  Quantities,  (70). — Least  Common  Multiple  of  Two  or  More  Quantities, 
(71). — Composition  of  the  Least  Common  Multiple,  (72). — Relation  of  Least 
Common  Multiple  to  Product  and  Greatest  Common  Measure,  (73). 

Rule  VIII.  To  Find  the  Least  Common  3Iultiple  of  Two  or  More 
Quantities,  (74). 


CHAPTER  IV. 

Fractions. — 41  ...  66. 

An  Algebraic  Fraction,  (75). — Of  a Quantity  with  a negative  Exponent, 
(76)  . — Two  Methods  of  representing  the  Quotient  when  the  Divisor  is  not  a 
Factor  of  the  Dividend,  (77). — How  any  Factor  may  be  transferred  from  tho 


viii 


ANALYSIS  OF  CONTENTS. 


Numerator  to  the  Denominator,  and  vice  versa,  (78). — Reciprocal  of  a 
Quantity,  (79). — Reciprocal  of  a Fraction,  (80). — Constant  Value  of  a 
Fraction,  (81). — When  a Fraction  is  positive,  and  when  negative,  (82). — 
Signification  of  + or  — prefixed  to  a Fraction,  (83). — Changes  of  Signs 
without  affecting  the  Value  of  a Fraction. — How  a Polynomial  may  be 
changed  from  Positive  to  Negative,  (84). — A Fraction  reduced  to  lower 
Terms — How  a binomial  Common  Measure  may  often  be  discovered,  (85). 

Rule  IX.  To  Reduce  a Fraction  to  its  Lowest  Terms,  (86). 

When  Two  or  more  Fractions  are  said  to  have  a Common  Denominator. 
—How  Fractions  may  be  reduced,  mentally,  to  a Common  Denominator, 
(87). 

Rule  X.  To  Reduce  Two  or  More  Fractions  to  a Common  Denominator,  (88). 

An  Integral  Quantity,  (89). — A Mixed  Quantity,  (90)  — An  Improper 
Fraction,  (91). 

Rule  XI.  To  Reduce  an  Integral  or  a Mixed  Quantity  to  an  Impropei 

Fraction , (92). 

Rule  XII.  To  Reduce  an  Improper  Fraction  to  an  Integral  or  a Mixed 

Quantity,  (93). 

By  what  means  the  Sum  of  Two  or  More  Fractions  is  found,  (94). 

Rule  XIII.  For  the  Addition  of  Fractions,  (95). 

By  what  means  the  Difference  of  Two  Fractions  is  found,  (96). 

Rule  XIV.  For  the  Subtraction  of  Fractions,  (97). 

Product  of  Two  or  more  Fractions,  (98). — Effect  of  Multiplying  by  t» 
Fraction,  (99). — Compound  Fractions,  (100). — Equivalent  of  Multiplying 
Two  or  more  Fractions  together,  (101). 

Rule  XV.  For  the  Multiplication  of  Fractions,  (102). 

How  a Fraction  is  Multiplied  by  its  own  Denominator,  and  what  Can- 
ecllations  may  be  made  in  the  Multiplication  of  Fractions,  (103). — Quotient 
of  Two  Fractions,  (104)  — Complex  or  Mixed  Fractions,  (105). 

Rule  XVI.  For  the  Division  of  Fractions,  (106). 


ANALYSIS  OF  CONTENTS. 


ix 


CHAPTER  Y. 

Simple  Equations. — 67  . . . 90. 

An  Equation — The  First  Member — The  Second  Member,  (107). — For 
what  purposes  Equations  are  employed — How  applied  to  the  Solution  of 
Questions,  (108). — The  Solution  of  an  Equation — Verification  of  the  Value 
found  for  the  Unknown  Quantity,  (109). — A Simple  Equation — A Quadratic 
Equation — A Cubic  Equation,  (110). — A Numerical  Equation — A Literal 
Equation — An  Identical  Equation,  (111). — Transformation  of  an  Equation, 
(112). — An  Axiom — Axiom  first,  second , &c.,  (113). — How  the  Value  of  the 
Unknown  Quantity  is  found — Transformations  necessary,  (114). — How  to 
clear  an  equation  of  Fractions — How  by  means  of  Least  Common  Multiple 
— Advantage  of  this  Method,  (115). — How  any  term  may  be  Transposed  from 
one  Side  of  an  Equation  to  the  other,  (116). — Change  of  the  Signs  in  an 
Equation,  (117). 

Rule  XVII.  For  the  Solution  of  a Simple  Equation  containing  but  one 
unknown  quantity , (118.) 

A Problem,  and  in  what  its  Solution  consists — General  Method  of  form- 
ing the  Equation  of  a Problem,  (119). — Solution  of  Problems  with  Two  or 
more  Unknown  Quantities — Independent  Equations,  (120). — General  Method 
of  solving  Two  Equations,  (121). — Elimination  by  Addition  or  Subtraction, 
(122). — Elimination  by  Substitution,  (123). — Elimination  by  Comparison, 
(124). — Solution  of  Three  Equations — Of  Four  Equations,  (125). — Of 
Problems  in  which  there  are  Three  or  more  Required  Quantities,  (126). 


CHAPTER  YI. 

Ratio. — Proportion. — Variation. — 91. ...  110. 

Ratio  of  one  Quantity  to  another,  (127). — Sign  of  Ratio,  (128). — How 
the  value  of  a Ratio  may  be  represented,  (129). — Direct  and  Inverse  Ratio, 
(130). — Compound  Ratio,  (131)  — Ratio  of  the  first  to  the  last  of  any  Num- 
ber of  Quantities,  (132). — Duplicate  and  Triplicate  Ratios,  (133). — Equi- 
multiples and  Equisubmultiples,  (134). — Ratio  of  Equimultiples  and  Equi- 
submultiples,  (135). — Proportion,  (136). — Four  Quantities  in  Proportion, 
(137).— Three  quantities  in  Proportion,  (138). — Direct  and  Inverse  Propor- 


X 


ANALYSIS  OF  CONTENTS. 


tion,  (139). — Sign  of  Proportion,  (140). — Inverse  Converted  into  Direct  Pro 
portion,  (141). — Variation — Variation  direct — Variation  inverse,  (142).— 
Product  of  Two  Quantities  varying  inversely  with  each  other,  (143). — Varia- 
tion, an  Abbreviated  Proportion,  (144). — A Theorem — A Corollary,  (145).— 
Ratio  of  two  Fractions  having  a common  Term,  (146). — How  the  value  of  a 
Fraction  varies , (147). — Corresponding  F.qualties  and  Inequalities  between 
the  Antecedents  and  Consequents  of  a Proportion,  (148). — A Proportion  con- 
verted into  an  Equation,  (149). — A Fourth  Proportional,  how  found,  (150). 
— Product  of  the  Extremes  when  Three 'Quantities  are  in  Proportion,  (151J. 
— Mean  Proportional,  how  found,  (152). — An  Equation  converted  into  a Pro- 
portion, (153). — R.at.io  of  the  first  to  the  third  of  Three  Proportional  Quanti- 
ties, (154). — Proportion  by  Inversion,  (155). — Proportion  by  Alternation, 
(156).  What  Multiplications  may  be  made  in  a Proportion,  (157). — What 
Divisions  may  be  made  in  a Proportion,  (158). — Proportion  by  Composition, 
(159). — Proportion  by  Division,  (160). — Proportion  between  the  Sum  of  two 
or  more  Antecedents  and  that  of  their  Consequents,  (161). — A Proportion 
derived  from  Two  other  Proportions  in  which  there  are  common  Terms, 
(162). — Proportion  between  the  Sums  and  Differences  of  the  Antecedents 
and  Consequents,  (163). — Products  of  the  Corresponding  Terms  of  Two  or 
more  Proportions,  (164).— Proportion  between  the  Powers  or  Roots  of  Pro- 
portional Quantities,  (165). — Substitution  of  Factors  in  a Proportion,  (166). 
— General  Solution  of  a Problem,  (167). — Two  Numbers  found  from  their 
Sum  and  Difference,  ( 1 68) . — An  Algebraic  Formula,  (169). — Of  a Propor- 
tion occurring  in  the  Solution  of  a Problem,  (170). — Percentage — Ratio  of 
Percentage — Basis  of  Percentage,  (171). — Amount  of  Percentage,  how  found, 
(172). — Interest — The  Principal — The  Amount,  (173). — Amount  of  Interest, 
how  found,  (174). 


CHAPTER  VII. 

Arithmetical,  Harmonical,  and  Geometrical  Progression. — 1 11...  120. 

An  Arithmetical  Progression,  (175). — Last  Term  of  an  Arithmetical 
Progression,  equal  to  what,  (176). — Common  Difference  of  the  Terms.  (177). 
— Sum  of  the  two  Extremes,  (178). — Arithmetical  Mean,  (179). — Sum  of 
all  the  Terms,  (180). — Formulas  in  Arithmetical  Progression,  (181). — An 
Harmonical  Progression,  (182). — An  Harmonical  converted  into  an  Arith- 


ANALYSIS  OF  CONTENTS. 


XI 


metical  Progression,  (183). — Harmonical  Mean,  (184). — A Geometerical  Pro- 
gression, (185). — Last  Term  of  a Geomitrical  Progression,  (186). — Power 
of  the  ratio  found  from  the  First  and  last  Terms,  (187). — Product  of  the  two 
Extremes,  (188). — Geometrical  Mean.  (189).— Sum  of  all  the  Terms,  (190) 
— Sum  of  an  infinite  number  of  Decreasing  Terms,  (191). — Formulas  in  Ge 
ometrical  Progression,  (192). 


CHAPTER  VIII. 

Permutations  and  Combinations. — Involution. — Binomial. — Theorem. 

— Evolution. — 121. . . . 146. 

Permutations,  (193). — Number  of  Permutations,  how  found,  (194). — 
Combinations,  (195). — Number  of  Combinations,  how  found,  (196). — Invo- 
lution, (197). — A Higher  Power  found  from  lower  Powers  of  thesame  Quan- 
tity. (198.) — Powers  of  unity,  (199). — Powers  of  Monomials,  how  found, 
(200). — Powers  of  Fractions,  how  found — Powers  of  a mixed  Quantity,  (201). 
Sign  to  be  Prefixed  to  a Power,  (202). — Powers  of  Binomials,  or  of  any  Po- 
lynomials, (203). — Binomial  Theorem — Exponents  in  any  Power  of  (a  fib) 
— Coefficients — Signs , (204). — Formula  for  the  development  of  (a fib  n) — At 
what  Term  the  development  will  terminate , (205). — Evolution — Extract- 
ing the  Square  Root,  in  what  it  consists — The  Cube  Root,  (206). — Roots  of 
unity,  (207). — Roots  of  Monomials,  how  found,  (208). — Roots  of  Fractions, 
how  found — Roots  of  a mixed  Quantity,  (209). — Of  a Root  whose  Exponent 
is  resolvable  into  two  Factors,  (210). — What  denoted  by  the  Numerator  and 
Denominator  of  a Fractional  Exponent,  (211). — Root  of  a power  of  a Quantity, 
(212). — Equivalent  Exponents.  (213). — Sign  to  be  Prefixed  to  an  odd  Root 
of  a Quantity,  (214). — Sign  to  be  Prefixed  to  an  even  Root,  (215). — Of  an 
even  Root  of  a negative  Quantity,  (216). — How  the  Roots  of  Polynomials 
may  be  discovered,  (217). 

Rule  XYIII.  To  Extract  the  Square  Root  of  a Polynomial , (218.) 

Principle  for  determining  the  Number  of  Figures  in  the  Square  Root  of  a 
Number.  (219). — Square  of  any  two  Parts  into  which  a number  may  be  di- 
vided.  (220). — Periods  to  be  formed  in  Extracting  the  Square  Root  of  a Deci- 
linal  Fraction— Why  the  last  Period  must  be  complete — Number  of  Decimal 
Figures  to  be  made  in  the  Root,  (221). 
fi 


jrii 


ANALYSIS  OF  CONTENTS. 


Rule  XIX.  To  Extract  the  Cube  Root  of  a Polynomial,  (222.) 

Principle  for  determining  the  Number  of  Figures  in  the  Cube  Root  of  a 
Number,  (223). — Cube  of  any  two  Parts  into  which  a Number  may  be  di- 
vided, (224). 

Rule  XX.  To  Extract  the  Cube  Root  of  a Number , (225). 

Periods  to  be  formed  in  extracting  the  Cube  Root  of  a Decimal  Fraction 
• — Why  the  last  period  must  be  Complete — Number  of  decimal  Figures  to  be 
made  in  the  Root, (226). — How  any  Root  whatever  of  a Polynomial  might  be 
extracted,  (227). 

Rule  XXI.  To  Extract  any  Root  of  a Polynomial , (228). 


CHAPTER  IX. 

Irrational  or  Surd  Quantities. — Imaginary  Quantities. — 147.  . .166. 

Perfect  and  Imperfect  Powers,  (229). — A Rational  Quantity — An  Irrationa. 
or  Surd  Quantity — Radical  Quantities,  (230). — Radical  Sign — How  this 
Sign  may  always  be  superseded,  (231). — Similar  and  Dissimilar  Surds. (232). 
How  a Rational  Quantity  may  be  expressed  under  the  Form  of  a Surd.  (233). 
Transfer  of  an  Exponent  between  Factors  and  Product,  (234). — To  what 
the  Exponent  of  a Quantity  may  be  changed,  (235). — Product  of  a Rational 
and  an  Irrational  Factor,  (236). — How  a Surd  maybe  simplified , (237). — 
How  a Fractional  may  be  reduced  ti>  an  Integral  Surd,  (238). — Surds  of 
Different  Roots  reduced  to  the  Same  Root,  (239). — How  to  find  the  Sum  or 
Difference  of  Similar  Surds — of  Dissimilar  Surds,  (240). — Howto  find  the 
Product  or  Quotient  of  Surds  of  the  same  root — of  different  Roots — of  any 
two  Roots  of  the  Same  Quantity.  (241). — Expediency  of  rationalizing  a Surd 
Divisor  or  Denominator,  (242). — How  a Monomial  Surd  may  be  made  to 
produce  a Rational  Quantity — a binomial  Surd — a trinomial  Surd,  (243). 
. — Of  the  Powers  and  Roots  of  Irrational  Quantities,  (244). — Of  the  Square 
Root  of  a Numerical  Binomial  of  the  form  a±y/b  (245). — Of  Imaginary 
Quantities — From  what  an  Imaginary  Quantity  results. (246). — Of  the  Cal- 
culus of  Imaginary  Quantities — By  what  means  the  Sign  affecting  the  Pro- 
duct of  two  Imaginaries  may  be  determined,  (247). — Resolution  of  an 
Imaginary  Quantity,  (248). — Product  of  two  Imaginary  Square  Roots,  1 2 4 9 ) 


ANALYSIS  OF  CONTENTS. 


£111 


CHAPTER  X. 

Quadratic  and  Other  Equatations. — 167  . ..  204. 

A Quadratic  Equation — A Cubic  Equation — A Biquadratic  Equation, 
(250). — A Pure  Equation — An  Affected  and  a Complete  Equation,  (251). — 
A Root  of  an  Equation,  (252). — Principle  for  determining  a Divisor  of  an 
Equation,  (253). — Principle  for  determining  a Root  of  an  Equation,  (254). 
Number  of  Roots  of  an  Equation — Whether  the  several  Roots  of  an  Eqau- 
tion  are  necessarily  unequal,  (255). 

Rule  XXII.  For  the  Solution  of  a Pure  Equation , (256). 

How  a Surd  Quantity  in  an  Equation  maybe  rationalized,  (257). — How 
two  Surds  in  an  Equation  may  be  rationalized,  (258). — Of  an  Equation  con- 
taining a Fraction  whose  terms  are  both  irrational , (259-). 

Rule  XXIII.  For  the  Solution  of  a Complete  Equation  of  the  Second  Degree , 

(260). 

How  any  Binomial  of  the  form  ax2-\-bx  maybe  made  a Perfect  Square  (261). 

Rule  XXIV.  To  Reduce  an  Equation  of  the  form  ax2-\-bx  to  a Simple  Equa~ 

tion,  (262). 

Of  Equations  having  a Quadratic  Form  with  reference  to  a Power  or 
Root  of  the  Uuknown  Quantity,  (263). — Of  the  Solution  of  two  Equations, 
one  or  both  of  the  Second  or  a Higher  Degree,  Containing  two  Unknown 
Quantities,  (264).— Solutions  by  means  of  an  Auxiliary  Unknown  Quantity, 
(265). — Solutions  by  means  of  two  Auxiliary  Unknown  Quantities,  (266). 
— General  Law  of  the  Coefficients  of  Equations,  (267). — Determination  of 
the  Integral  Roots  of  Equations,  (268). — Solution  of  Equations  by  Ap 
proximation , (269). — General  Method  of  Elimination, (270). 


CHAPTER  XI. 

General  Description  of  Problems. — Miscellaneous  Problems. — 205  . 

217. 

A Determinate  Problem — How  a Determinate  Problem  is  represented,  (270). — 
An  Indeterminate  Problem— How  represented,  (271). — An  Impossible  Problem — 
How  represented,  (272). — Greatest  Product  of  any  two  Parts  into  which  any 
Quantity  may  be  divided,  (273). — Signification  of  the  Different  Forms  under  which 
the  value  of  the  Unknown  Quantity  may  be  found  in  an  Equation,  (274). — Inequa- 
tions,^ 275). — For  what  purpose  Inequations  are  employed,  (276). — Inequalities  iu 
the  same,  and  in  a contrary  sense,  (277). — Inequalities  between  negative  quantities, 
(278). — Transformation  of  Inequations,  (279). 


' 


ALGEBRA 


CHAPTER  I. 

PRELIMINARY  DEFINITIONS  AND  EXERCISES 

Science  and  Art. 

(1.)  Science  is  knowledge  reduced  to  a system. — Art  is  know- 
ledge applied  to  practical  purposes. 

The  Rules  of  Art  are  founded  on  the  Principles  of  Science. 

Numbers. — Quantity. 

(2.)  A unit  is  any  thing  regarded  simply  as  one ; and  numbers 
are  repetitions  of  a unit. 

Thus  the  numbers  two,  three,  Ac.,  are  repetitions  of  the  unit  one. 

(3.)  Quantity  is  any  thing  which  admits  of  being  measured. 

Thus  a line  is  a quantity,  and  we  express  its  measure  in  saying  it 
is  so  many  feet  or  inches  long.  Time,  'weight,  and  distance  are  also 
quantities. 

Numbers  are  quantities ; for  every  number  expresses  the  measure 
of  itself  in  units;  and  numbers  are  used  to  express  the  measures  of  all 
other  quantities.  Thus  we  express  the  measure  of  Time  by  a number 
of  days,  hours,  Ac 


Mathematics. 

(4.)  Mathematics  is  the  science  of  quantity.  Its  most  general 
divisions  are  Arithmetic  and  Geometry. 


2 


PRELIMINARY  DEFINITIONS  AND  EXERCISES. 


Arithmetic  is  the  science  of  numbers ; or,  when  practically  ap 
plied,  the  art  of  Calculation. 

Geometry  is  the  science  which  treats  of  Extension — in  length 
breadth , and  height,  depth,  or  thickness. 


(5.)  Algebra  is  a method  of  investigating  the  relations  of  both 
Arithmetical  and  Geometrical  quantities,  by  signs  or  symbols 

Symbols  of  Quantities. 

(6.)  Quantities  are  represented  in  Algebra  by  letters. — known 
quantities  usually  by  the  first , and  unknown  or  required  quantities  by 
the  last  letters  of  the  Alphabet. 

Thus  the  quantity  a or  b,  that  is,  the  quantity  represented  by  a oi 
b,  will  generally  he  understood  as  known  in  value ; while  the  quan- 
tity x or  y will  be  unknown  or  required. 

Quantities  represented  hv  letters  are  called  literal  quantities,  in 
contradistinction  to  numbers  or  numerical  quantities. 

Symbols  of  Operations. 

(7.)  The  sign  + plus  prefixed  to  a quantity,  denotes  that  the  quan 
tity  is  to  be  added,  or  taken  additively. 

Thus  a-\-b,  a plus  b,  denotes  that  the  quantity  b is  to  he  added  t< 
the  quantity  a. 

(3.)  The  sign  — minus  prefixed  to  a quantity,  denotes  that  the 
quantity  is  to  he  subtracted,  or  taken  subtract ively. 

Thus  a — b,  a minus  b,  denotes  that  the  quantity  b is  to  he  sub- 
tracted from  the  quantity  a. 

(9.)  The  sign  X into  between  two  quantities,  denotes  that  the 
two  quantities  are  to  he  multiplied  together. 

Tims  ax 5,  a into  b,  denotes  that  the  two  quantities  a and  b are  to 
be  multiplied  together. 

A point  (.)  between  two  literal  quantities,  or  a numeri  al  and  a 
literal  quantity,  also  denotes  that  the  two  quantities  are  to  be  multi- 
plied together. 

a.b  denotes  a into  b,  and  3. a denotes  3 into  a.  or  3 times  a. 

The  Product  of  several  numbers  maybe  denoted  b-  points  or  wven 
them;  thus  1.2.3  denotes  1 into  2 into  3,  the  same  as  1x2x3. 


PRELIMINARY  DEFINITIONS  AND  EXERCISES. 


3 


Quantities  in  juxtaposition,  without  any  sign  between  them,  are 
to  he  multiplied  together.  Thus  ab  denotes  a and  b multiplied  toge- 
ther ; and  axy  denotes  a,  x,  and  y multiplied  together. 


(10.)  The  sign  -4-  by  between  two  quantities,  denotes  that  the 
quantity  before  the  sign  is  to  he  divided  by  the  one  after  it. 

Thus  a -4-5,  a by  b,  denotes  that  a is  to  be  divided  by  b. 


Division  is  also  denoted  by  placing  the  dividend  over  the  divisor , 
with  a line  between  them,  after  the  manner  of  a Fraction ; 


- denotes  a divided  by  b,  the  same  as  a- 


-b. 


An  integral  quantity , in  Algebra,  is  one  which  does  not  express 
any  operation  in  division,  whatever  may  be  the  numerical  values 
which  the  letters  represent. 


(11.)  A parenthesis  ( ) enclosing  an  algebraic  expression,  or  a 

vinculum  drawn  over  it,  connects  the  value  of  that  expression 

with  the  sign  which  immediately  precedes  or  follows  it. 

Thus  (<2  + 5).c,  or  (<2  + 5)c,  a plus  5 in  a parenthesis  into  c,  denotes 
that  the  sum  of  a and  5 is  to  be  multiplied  into  c. 

The  same  thing  would  be  denoted  by  a+bxc,  a plus  5 under  a 
vinculum  into  c. — In  af-bc,  only  5 would  he  multiplied  into  c. 

The  vinculum,  and  the  expression  affected  by  it,  are  sometimes 
set  vertically. 

Thus  a x is  equivalent  to  (<2  + 5 — c)  x 

+ £*  

— c or  af-b—cxx. 


PRELIMINARY  EXERCISES. 

In  the  elementary  oral  Exercises  which  are  occasionally  inserted, 
the  Student  should  write  down  the  quantities  as  they  are  read  to  him. 

Suppose  the  letters  a,  b,  c to  represent  the  numbers  3.  4 5, 
respectively ; then  what  is  the  numerical  value  of  <2  + 5 — c ? 

What  is  the  value  of  <z5+c?  Of  abc — 5c+<2  ? 

What  is  the  value  of  ac-t-bl  Of  ab  +5  — ac  ? 

What  is  the  value  of  bc-^a  ? Of  abc+ac—bc  ? 

What  is  the  value  of  (<2  + 5)  c,  a plus  5 in  a parenthesis  into  c ? 

What  is  the  value  of  («5  + c)  a ? Of  ( a +5  + c)  c ? 

What  is  the  value  of  (a  + 5)-4-c?  Of  (c/5  + 5 — c)  <2  ? 

What  is  the  value  of  (5c  — «)4-5?  Of  (be — « + 6)5? 

The  Teacher  may  propose  other  Exercises  of  the  same  nature,  should 

he  deem  it  necessary. 


4 


PRELIMINARY  DEFINITIONS  AND  EXERCISES. 


Factors. — Constant  Product. 

(12.)  Two  or  more  quantities  multiplied  together,  are  called  the 
factors  of  their  product ; and  the  Product  is  the  same  in  value , in 
whatever  order  its  factors  are  taken. 

Thus  a and  x are  the  factors  of  the  product  ax  ; and  this  product 
is  the  same  in  value  as  xa.  So  a,  b,  and  c are  the  factors  of  the  product 
abc,  or  acb , or  bca,  See. 

It  is  most  convenient  to  set  literal  factors  according  to  the  order  of 
the  same  letters  in  the  Alphabet ; thus  ax ; abc. 

To  understand  why  the  product  ax  is  equal  to  xa , consider  a and  x 
as  representing  numbers,  and  that  the  product  of  two  numbers  is  the 
same,  when  either  of  them  is  made  the  multiplier. 

For  example,  25  times  7 is  equal  to  7 times  25.  For  25  times  7 
must  be  7 times  as  many  as  25  times  1,  which  is  25 ; that  is,  25  times 
7 is  equal  to  7 times  25. 

[CP”  Prove  that  14  times  9 is  equal  to  9 times  14. 

Prove  that  31  times  1 1 is  equal  to  11  times  31. 

Prove  that  23  times  15  is  equal  to  15  times  23. 

Prove  that  47  times  18  is  equal  to  18  times  47. 


Pouters  and  Roots. 

(13.)  The  first  pouter  of  a quantity  is  the  quantity  itself ; thus 
the  first  power  of  5 is  5,  and  the  first  power  of  a is  a. 

Tire  second  power,  or  square,  of  a quantity,  is  the  product  of  the 
quantity  multiplied  into  itself.  Thus  the  second  power,  or  square,  of 
5 is  5 x 5,  which  is  25  ; and  the  second  power  of  a is  aa. 

The  third  power,  or  cube,  of  a quantity  is  the  product  of  the  quan- 
tity multiplied  into  its  second  power,  or  square.  Thus  the  third  power, 
or  cube  of  5 is  5 X 5 X 5,  which  is  125  ; and  the  third  power  of  a is  aaa. 

[CP"  What  is  meant  by  the  fourth  power  of  a quantity  ? What 
is  meant  by  the  fifth  power  of  a quautity  ? By  the  seventh  power  of 
a quantity  ? 

What  is  the  square  of  3 ? The  cube  of  4 ? The  fourth  power  of 
2 ? The  square  of  7 ? The  cube  of  6 ? The  fourth  power  of  10  ? 


PRELIMINARY  DEFINITIONS  AND  EXERCISES. 


5 


(14.)  The  second  root , or  square  root,  of  a quantity,  is  that  quan- 
tity whose  square  is  equal  to  the  given  quantity.  Thus  the  square 
root  of  9 is  3 ; and  the  square  root  of  aa  is  a. 

The  third  root , or  cube  root,  of  a quantity,  is  that  quantity  whose 
| third  'power,  or  cube,  is  equal  to  the  given  quantity.  Thus  the  cube 
root  of  8 is  2 ; and  the  cube  root  of  aaa  is  a. 

What  is  meant  by  the  fourth  root  of  a quantity  ? What  is 
meant  by  the  fifth,  root  of  a quantity  ? By  the  ninth  root  of  a quantity  ? 

What  is  the  square  root  of  16  ? The  cube  root  of  27  ? The  fourth 
root  of  16  ? The  square  root  of  81  ? The  cube  root  of  1000  ? 

What  is  the  square  of  the  square  root  of  4 ? The  cube  of  the  cube 
root  of  125  ? The  square  of  the  cube  root  of  64?  The  cube  of  the 
square  root  of  1 6 ? 


Coefficients  and  Exponents. 

(15.)  The  coefficient  of  a quantity  is  any  multiplier  prefixed  to 
that  quantity. — In  a more  general  sense,  the  coefficient  of  a quantity 
is  any  factor  forming  a product  with  that  quantity. 

Thus,  in  3 a,  3 is  the  coefficient  of  a,  and  denotes  3 times  a.  In 
5ax,  5 is  the  coefficient,  denoting  5 times  ax.  In  \x,  ^ is  the  coefficient 
of  x,  and  denotes  one-half  of  x. 

When  no  numerical  coefficient  is  prefixed,  a unit  is  always  to  he 
understood.  Thus  a is  la,  once  a,  and  ax  is  lax,  once  ax. 

(16.)  The  exponent  of  a quantity  is  an  integer  annexed  to  it,  to 
denote  a poiver,  or  a fraction  annexed  to  denote  a root,  of  that  quantity. 

Thus  a2,  a with  exponent  2,  denotes  the  second  power,  or  square 
of  a;  x3  denotes  the  third  power,  or  cube,  of  x ; and  so  on. 

The  fractional  exponents  4,  -J,  and  so  on,  denote,  respectively 
the  square  root,  cube  root,  &c.,  of  the  quantity  to  which  they  are 
annexed. 

i . l 

a 2,  a with  exponent  denotes  the  square  root  of  a;  Xs  denotes 
the  cube  root  of  x ; and  so  on. 

When  no  exponent  is  annexed  to  a quantity,  a unit  is  always  to  he 
understood.  Thus  a is  a1,  the  first  power  of  a.  (13.) 


6 


PRELIMINARY  DEFINITIONS  AND  EXERCISES. 


An  exponent  is  assigned  to  the  'product  of  two  or  more  fact;  rs  by- 
affecting  such  product  with  a parenthesis,  or  a vinculum,  and  the 
exponent. 

Thus  (ax)2  or  ax2,  ax  in  a parenthesis,  or  under  a vinculum,  with 
exponent  2,  denotes  the  square  of  the  product  ax ; whereas  ax2  denotes 
a into  the  square  of  x. 

(17.)  An  integral  coefficient  indicates  the  repeated  addition  of  a 
quantity  to  itself ; while  an  integral  exponent  indicates  the  repeated 
multiplication  of  a quantity  into  itself. 

Thus  3a,  3 times  a,  is  equivalent  to  a-\-a-\-a  ; while 
a3,  the  third  power  of  a,  is  equivalent  to  aaa. 

Coefficients  and  exponents  are  thus  employed  to  abbreviate  the 
language  of  Algebra. 

□GP”  What  is  the  equivalent,  in  Addition,  of  2x  ? Of  3 ax  ? Of 
Zabc  ? Of  5a 2 ? Of  4 ay2  ? Of  3 axy  ? 

What  is  the  equivalent,  in  Multiplication,  of  a2  ? Of  x3  ? Of 
ax2  ? Of  a2x  ? Of  abx 3 ? Of  ac2x 2 ? 

Allowing  the  value  of  a to  be  4,  what  is  the  value  of  a2  ? Of  a-  ? 
Of  a3  ? Of  5a2  ? Of  10a-  ? Ofa'-a3?  Of  (4a)- ? 

i i 

Allowing  a to  be  4,  and  b 9,  what  is  the  value  of  ab'-  ? Of  a-b  ? 
Of  a2$  ? Of  (abf2  ? Of  ah%  ? Of  \ab  ? Of  \(abf  ? 


Similar  and  Dissimilar  Quantities. 

(18.)  Similar  quantities  are  such  as  have  all  the  literal  facto?  s, 
with  their  respective  exponents,  the  same  in  each;  otherwise,  the 
quantities  are  dissimilar. 

Thus  the  two  quantities  2 ax2  and  5ax2  are  similar  ; while  oax2 
and  4 a2x  are  dissimilar,  the  literal  factors  not  having  the  same  expo- 
nents in  each. 

[GP”  Are  the  two  quantities  ab  and  Sab  similar,  or  dissi?nilar  ? 
Are  4a  and  3a;  similar,  or  dissimilar  ? Are  ay  and  3 ya  similar,  or  dis- 
similar? 5aoy-  and  lafxl  abc2  and  5bc2  ? 2ab  and  2a62  ? 3 axy2 
and  ay2x  ? 2 be2  and  35c-  ? 

Give  an  example  of  three  similar  quantities. — Give  an  example 
of  three  dissimilar  quantities. — Another  example  of  three  similar  quan- 
tities.— Another  example  of  three  dissimilar  quantities. 


PRELIMINARY  DEFINITIONS  AND  EXERCISES. 


7 


Monomials  and  Polynomials. 

(19.)  An  algebraic  monomial  is  a symbol  of  quantity  not  com- 
posed of  parts  connected  by  tbe  sign  -f-  or  — . 

Thus  3a,  5ax,  2x2,  and  \abc3  are  monomials. 

(20.)  An  algebraic  •polynomial  consists  of  two  or  more  monomials 
connected  by  the  sign  -f-  or  — ; and  such  monomials  are  called  the 
terms  of  the  polynomial. 

Thus  5a2-\-bx,  and  ax2  +36  — 5c2  are  polynomials. 

A polynomial  composed  of  two  terms  is  called,  more  definitely,  a 
binomial,  and  one  composed  of  three  terms,  a trinomial. 

(21.)  The  value  of  a polynomial  is  not  affected  by  changing  the 
order  of  its  terms,  without  changing  the  sign  prefixed  to  any  term. 

Thus  a-\-b — c is  equivalent  to  a — c-\-b ; for  the  result  will  evidently 
be  the  same,  whether  b be  first  added  to  a,  and  c then  subtracted  ; or 
c be  first  subtracted,  and  b afterwards  added. 

(22.)  A polynomial  is  arranged  according  to  the  'powers  of  one  of 
its  letters,  when  the  exponents  of  that  letter  increase,  or  decrease,  con- 
tinually in  the  successive  terms. 

Thus  the  polynomial  3a3  + 4a 2 x — 5ab,  is  arranged  according  to 
the  descending  powers  of  a,  since  the  exponents  of  a decrease  contin- 
ually in  the  successive  terms. 

The  letter, — as  a in  this  example, — according  to  which  the  poly- 
nomial is  arranged,  is  called  the  letter  of  arrangement. 

(23.)  A polynomial  is  said  to  be  homogeneous,  when  the  sum  of 
the  exponents  of  the  literal  factors,  is  the  same  in  each  of  its  terms. 

Thus  a3-\-2ax2 — bey  is  homogeneous,  since  the  sum  of  the  expo- 
nents of  the  literal  factors  is  the  same,  namely  3,  in  each  term. 

Each  one  of  the  literal  factors  composing  a term,  is  called  a,  dimen- 
sion of  that  term ; and  the  degree  of  any  term  is  the  ordinal  of  the 
number  of  its  literal  factors  or  dimensions. 

Thus  4a2a:  contains  three  literal  factors,  aax,  and  is  therefore  of 
three  dimensions,  or  of  the  third  degree. 

CF3*  Arrange  the  polynomial  2x-\-3a2x2 — 4 ax3+a3y2  according 
to  the  descending  powers  of  a. — Arrange  it  according  to  the  ascending 
powers  of  a. — Which  of  the  terms  of  this  polynomial  are  homogeneous  ? 
Tell  the  number  of  dimensions  in  each  term. — Of  what  degree  is  each 
term  ? 


8 


PRELIMINARY  DEFINITIONS  AND  EXERCISES. 


Positive  and  Negative  Quantities. 

(24.)  A 'positive  quantity  is  one  which  enters  additivcly,  and  \ 
negative  quantity  is  one  which  enters  subtractively , into  a calculation. 

A positive  quantity  is  denoted  by  the  sign  + > and  a negative  quan 
tit.y  by  the  sign  — , prefixed  to  it. 

In  the  polynomial  a-\-b — c,  the  quantity  b is  positive,  while  c ia 
negative. 

A quantity  with  neither  + nor  — prefixed  to  it,  is  understood  tc 
be  positive. 

Thus  in  the  preceding  polynomial,  the  first  term  a is  understood  t$ 
be  +«  ; for  it  may  be  regarded  as  0 + a. 

(25.)  A negative  quantity  has  an  effect  contrary  to  that  of  ar 
equal  positive  quantity,  in  the  expression,  or  calculation,  into  which  it 
enters. 

This  is  evident  with  regard  to  the  addition  and  subtraction  of  the 
same  quantity ; the  effect  of  subtracting  it  is  contrary  to  that  of  add- 
ing it. 

The  effect  of  multiplying  by  a negative  quantity  is  contrary  to  that 
of  multiplying  by  an  equal  qjositive  quantity. 

For  example;  3a  X 2,  3a  multiplied  by  positive  2,  denotes  that  3a 
is  to  be  repeated  additively,  and  is  equivalent  to  3a  + 3a. 

But  3a  X — 2,  3a  multiplied  by  negative  2,  denotes  that  3a  is  to  be 
repeated  subtractively,  and  is  equivalent  to  — 3a  — 3a. 

The  same  thing  is  true  with  respect  to  dividing  by  a negative,  and 
an  equal  positive  quantity. 

The  positive  and  negative  signs  are  sometimes  used  to  distinguish 
quantities  as  estimated  in  contrary  directions  from  a given  point  or  line. 

Thus  if  Motion  in  one  direction  be  denoted  by  the  sign  +.  then 
motion  in  the  contrary  direction  will  be  denoted  by  the  sign  — . Ii 
North  Latitude  be  considered  as  positive,  South  Latitude  will  be 
negative. 


9 


CHAPTER  II. 

ADDITION— SUBTRACTION.— MULTIPLICATION.— DIVISION. 

ADDITION. 

(20  1 Algebraic  'Edition  consists  m finding  the  simplest  expres 
sion  foi  the  value  ot  two  or  more  quantities  connected  together  by  th« 
sign  4-  or  — , and  this  equivalent  expression  is  called  the  sum  of  tht 
quantities. 

The  simple  t expression  for  the  value  of  5a-f  3a,  5 times  a plus  3 
times  a,  is  8a ; then  8a  is  the  sum  of  5a  and  3a. 

The  simplest  expression  for  the  value  of  5a  — 3a,  5 times  a minus 
3 times  a,  is  2 a ; then  2 a is  the  sum  of  5a  and  — 3a. 

In  the  second  example,  observe  that  adding  — 3a  is  equivalent  to 
subtracting  3 a ; the  Adding  of  a negative  quantity  being  the  same  as 
the  Subtracting  of  an  equal  positive  quantity. 

Addition  of  Monomials. 

(27.)  Similar  terms  with  like  signs,  are  added  together  by  taking 
the  sum  of  their  coefficients , annexing  the  common  literal  factor , and 
prefixing  the  common  sign. 

Thus  3 ax-\-2ax  is  5ax ; just  as  3 cents  + 2 cents  is  5 cents. 

And  — 3a  — 2a  is  — 5a ; for  3a  to  be  subtracted  and  2a  to  be  sub- 
tracted make  5a  to  be  subtracted. 

HdP*  What  is  the  Sum  of  4a  and  3a  ? Of  ax  ar  1 5ax  ? Of  — ° 
and  —7xl  Of  5a2,  2a2,  and  4a2?  Of  — 2b,  —3 b,  and  — 5br> 

(28.)  Two  equal  similar  terms  with  contrary  signs,  when  added 
together,  mutually  cancel  each  other ; that  is,  their  sum  is  0. 

Thus  3a  — 3a  is  0 ; that  is,  0 is  the  simplest  express:-ev  for  the 
value  of  3a  — 3a,  c*--3  is  therefore  the  sum  of  the  two  terrr  . 

So  5a-\-x — 5a  .s  equal  to  x;  for  5a  and  - -5a  cancel  each  other, 
ar  x is  therefore  the  sum  of  the  three  given  terms 


10 


ADDITION. 


What  is  the  Sam  of  2 ax  and  — 2 ax,  that  is,  the  simplest  ex- 
pression for  the  value  of  2ax—  2ax1  What  is  the  Sum  of  76  and  \ 
— 76?  Of  a,  3x,  and  — a?  Of  3 y2,  — 3 y2 , and  5b2  ? Of  ax2,  5, 
and  — 5 ? Of  —abc,  -\-a2x,  and  abc  ? 

(29.)  Two  unequal  similar  terms  with  contrary  signs,  are  added 
together  by  taking  the  difference  of  their  coefficients,  annexing  the 
common  literal  factor,  and  prefixing  the  sign  of  the  greater  term. 

Thus  la— 3a  is  4 a;  just  as  7 vents  — 3 cents  is  4 cents;  that  is, 
the  sum  of  la  and  — 3 a is  4 a. 

And  3a— la  is  —4 a.  For  —la  is  equal  to  —3a  — 4a  (27);  then 
3a — la  is  equal  to  3a  — 3a  — 4a;  3a  and  — 3a  cancel  each  other  (2S), 
and  leave  —4a. 

KF5"  What  is  the  Sum  of  5ab  and  — 3a6  ? Of  3a2c  and  — a2c? 
Of  9 ax2  and  — 4aa;2  ? Of  — a2x  and  3a2z?  Of  5ac  and  — 5 oc  ? 

What  is  the  simplest  expression  for  the  value  of  2b  — 5b,  that  is, 
the  Sum  of  2b  and  — 5b  ? How  do  you  reason  in  finding  that  sum  ? 

What  is  the  Sum  of  3a2  and  — 8a2  ? and  how  do  you  reason  in 
finding  it?  What  is  the  Sum  of  ax2  and  — 2 ax2  ? and  how  do  you 
reason  in  finding  it?  What  is  the  Sum  of — 11a2  and  5a2?  and 
how  do  you  reason  in  finding  it  ? 

(30.)  The  Sum  of  two  or  more  dissimilar  terms  can  only  he  indi- 
cated by  arranging  them  in  a Polynomial,  so  that  each  term  shall  have 
its  given  sign  prefixed  to  it. 

Thus  the  Sum  of  ax  and  be 2 can  only  be  indicated  as  axfibc2  ; and 
the  Sum  of  ax  and  — be 2 as  ax — be2. 

What  is  the  Sum  of  3a  and  5b  ? Of  5a  and  — b ? Of  3a2  and 
2a:2  ? Of  ax  and  by  ? Of  —36  and  — 4 y ? 

What  is  the  Sum  of  36,  56,  and  6c?  Of  a2,  bx,  and  3a2  ? Of 
4c  and  —3i / ? Of  —ab,  3 xy,  and  —2c  ? Of  3a2,  46,  and  2c  ? Of 
5a,  — 5,  and  56  ? 

The  preceding  principles,  and  the  following  Rule,  provide  for  all 
the  cases  in  Algebraic  Addition. 


ADDITION. 


11 


RULE  I. 

(31  ) For  the  Addition  of  Algebraic  Quantities. 

1.  Find  the  positive  and  the  negative  Sum  of  similar  terms,  sepa 
rately,  (27),  and  then  add  together  the  similar  sums,  (28)  and  (29). 

2.  Connect  the  results  thus  found,  and  the  dissimilar  terms,  in  a 
Polynomial,  prefixing  to  each  term  its  proper  sign.  (30). 

EXAMPLE. 

To  add  together  4a2  +2bc— xy,  2a2—3bc+5y,  bc—a2  + 3xy,  and 
cy—2a2—5bc. 

4a2  + 2bc — xy 
2 a2  — 3 bc-\-5y 
— a2  + bcA^xy 
— 2 a2  — 55c+q/ 


3a2  —5bc-\-2xy+cy-\-5y 

The  sum  of  the  positive  terms  2a2  and  4a2  is  6a2,  and  the  sum  of 
the  similar  negative  terms  —2a2  and  —a2  is  — 3a2,  .(27).  Adding 
together  these  two  similar  sums,  6a2  and  —3a2,  the  residt  is  3a2,  (29). 

The  sum  of  the  positive  terms  be  and  2 be  is  35c,  and  the  sum  of  the 
similar  negative  terms  — 55c  and  — 35c  is  —85c.  Adding  together 
these  two  similar  sums,  35c  and  — 85c,  the  result  is  — 55c. 

The  sum  of  the  similar  terms  3 xy  and  —xy  is  2 xy. 

The  results  thus  found,  namely,  3a2,  — 55c,  and  2 xy,  and  the 
dissimilar  terms  cy  and  5y,  are  connected  in  a Polynomial. 

EXERCISES. 

1.  Add  together  a5  + 3c2 — 2x,  oab — c2  + 5.r,  5c2  -2ab-{-y,  and 
4a5— c2+x.  Ans.  6a5  + 6e2  + 4a;+?/. 

J 2.  Add  together  4a2 — 65  + 3,  3a2  + 65  — 7,  55  + a2— 5c,  and  a2  — 
5c+10.  Ans.  9a2  + 55  + 6— 25c. 

3.  Add  together  252+5c+a;,  352 — 25c  + ?/,  5c  — 352  + 5.r,  and 
52  + 5c+3y.  Ans.  352+5c+6x+4?/. 

4 Add  together  5a  — 653  + 3,  253 — 4a  — 7,  7a  — 53+c,  and  a+ 
352  + 2c+4.  Ans.  9a — 553  + 3c+352. 

5.  Add  together  a25  — 5c+zi/,  4a25  + 8c — xy,  a25  + c+3:c?/,  and 
b2+3c— 13.  Ans.  6a25+7c+322/+a2 — 13. 

B 


12 


ADDITION. 


6.  Add  together  a3  + 2a2  —35c,  2a3—  a2  Abe,  — a3 — a2+£2 , 

and  3a3  + 3a2  + 362.  Am.  5a3  + 3a2 -26c+462. 

7.  Add  together  2x+ab 2 —3y2,  2ab2  — 3x-\-iy2,  — ab2  + 5x—y2, 

and  5.c+a52 — y.  Aws.  6x-\-3ab2 — 3?/2 — ?/. 

8.  Add  together  4a — 352+ca;2,  b2 — 3a+3,  2a+362 — cx2 , and 

2 b2 — 2a  — 9.  A?is.  a + 362— 6. 

9.  Add  together  lb — 2 a2—xy,5a2 — 6b-\-3xy,  b— 3a2+4,  and 

a? — b — 3 xy.  Ans.  b-\-a2 — zy+4. 

10.  Add  together  3 b2 — 2as  + l3,  3a3  — 2 b2—5,  Aab-\-7a3  — 3, 

and  2 b2 — a3-\-ab.  Am.  3d2  + 7a3  + 5 + 5aA 

11.  Add  together  ax2 — 2 yfb,  2y-\-2ax2 — 3 b,  Aax2—y—b,  and 

2b—3ax2Ab.  A?is.  Aax2—y. 

12.  Add  together  2c2+a2+35c,  5c2 — 3a2 — 2 be,  c2  + 2a2 — be, 

and  bAbc— 3.  Am.  8 c2-\-bc-\rb — 3. 

13.  Add  together  abfa2c — 5,  3 ab — 3a2c+7,  2a2c—2ab  — 3, 

and  a5  + a2c+5.  Ans.  3a6+a2c+4. 

14.  Add  together  3 b*  — 2 a2x-\-b,  — b2-\-3a2x — 3 b,  b2 — a2x\-c, 

and  3 b2 -\-b — 3t-.  Ans.  6b2—  b — 2c. 

15.  Add  together  2a2  + 3 — ac,  3 a2 — 7+ac,  3ac— 5a2  + 9,  and 

3ac+4 — a2.  Ans  — a2  + 9 + 6ac. 

16.  Add  together  £2c+ 2 — y2,  y2 — 3b2c— 10,  2 b2c — 3 + 2 y2, 

and  b2 — ?/2  + 5.  Ans.  — 6+y2  + 62. 

17.  Add  together  a3+Sc  — £c,  2 a3 — £c+fc,  3a3  + 3ic — \c,  and 

a3  + 5c+c.  Ans.  7a3  + 46c+c. 

18.  Add  together  b3 — 3a2+fn2,  262  + 5a2+d-c2,  b2 — -^c+2, 

and  5b2 + 3a2 -2c-\-3y.  Ans.  63  + 5a2  + l-|c2  + S62  — 2^c+2+3y. 

(32.)  When  the  same  Polynomial  is  enclosed  in  two  or  more  paren- 
theses, (11),  this  polynomial  enters  into  a calculation  in  the  same  man- 
ner as  the  common  factor  of  similar  monomials. 

Thus  the  Sum  of  3(a  + b)  and  5(a  + b),  is  evidently  8(a+S), 

19.  Add  together  2(a  — 3Z>)  + c,  3(a— 3Z>)— 3c,  — 4(a— 35)  + c, 

and  5(a— 3&)  + 3c.  A?is.  6(a— 36)  + 2c. 

20.  Add  together  3a2  + 2(a— c2),  a2— 3(a— c2),  4a2+5(a—  c2) 

and  — 7a2  + (a— c2).  A?is.  a2  + 5(a— c2). 

21.  Add  together  5(a  + 5— c)  + 3&2,  3(a-\-b—c)  — 2b2,  2(a-\-b—c) 

and  3b2  + 2?/,  -\-b2 . Ans.  10 (a  + i — c)-\-5b2 -\-2y. 

22.  Add  together  2aZ>2  + 3a(£  + c2),  ab2  — 2a(&+c2),  a(b+c2) 

and  — ab2  — 3a(5  + c2).  Ans.  2ab2  — a(b+c2). 

23.  Add  together  5 + -|{c— d+m)-\-2b,  1 +f(c— d+m)—  b + 2, 

and  2 {c—d+m)+%b.  Ans.  8 + 3i(c— cf+w)  + 1^-6. 


SUBTRACTION 


13 


SUBTRACTION. 

(33.)  Algebraic  subtraction  consists  in  finding  the  difference 
between  the  two  quantities  ; that  is,  in  finding  what  quantity  added 
to  the  quantity  subtracted , will  produce  that  from  which  the  subtrac- 
tion is  made. 

Thus  the  difference  found  by  subtracting  3 a from  5a  is  2a,  because 
2 a added  to  3a  produces  5a,  (27). 

But  the  difference  found  by  subtracting  — 3 a from  5a  is  8a ; be- 
cause 8 a must  be  added  to  — 3a  to  produce  5a,  (29). 

In  the  second  example,  observe  that  subtracting  — 3a  has  an  effect 
contrary  to  that  of  subtracting  3a  in  the  first  example,  (25)  ; the  Sub- 
tracting of  a negative  quantity  being  the  same  as  the  Adding  of  an 
equal  positive  quantity. 

Subtraction  of  Monomials. 

(34.)  A monomial  is  subtracted  from  another  quantity,  by  changing 
the  sign  of  the  monomial,  and  then  adding  it,  algebraically,  to  the 
other  quantity. 

Thus  5a  subtracted  from  2a  gives  — 3a,  because  — 3a  must  be 
added  to  5a,  to  produce  2a,  (29)  ; and  this  difference,  — 3a,  will  be 
found  by  changing  5a  to  — oa,  and  adding  the  — 5a  to  the  2. a 

[TP*  What  difference  will  be  found  by  subtracting  5a  from  9a  ? 
and  why  is  that  difference  true?  By  subtracting  9a  from  5a? 

What  difference  will  be  found  by  subtracting  3 ab  from  4 ab  1 and 
why  is  that  difference  true?  By  subtracting  — 5x  from  — 10a;?  and 
why  is  that  difference  true  ? By  subtracting  lax  from  — 3 ax. 

(35.)  When  the  two  quantities  are  dissimilar,  a monomial  is  sub- 
tracted by  changing  its  sign,  and  then  connecting  it  with  the  other 
quantity,  in  a Polynomial. 

Thus  2x  subtracted  from  3a  gives  3a  — 2x  ; and  — 2x  subtracted 
from  3a  gives  3a  + 2cc,  (34),  (30). 

[TP  What  difference  will  be  found  by  subtracting  2a  from  3a; ? 
By  subtracting  —35  from  5axl  By  subtracting  ax  from  105?  By 
subtracting  4 y from  —9a:  2 ? 


14 


SUBTRACTION. 


From  the  preceding  we  derive 

RULE  II. 

(36.)  For  the  Subtraction  of  Algebraic  Quantities. 

Change  the  sign  of  each  term  to  be  subtracted,  + to  — and  — to 
+ , or  conceive  these  signs  to  be  changed,  and  then  proceed  as  in  Alge- 
braic Addition. 

EXAMPLE. 

To  subtract  a2  + 2a£  — 3je — z from  3a2—  5ab-\-5x — y. 

3 a2  — 5 ab-\-5x — y 
a2  -\-2ab — 3x — z 


2 a2 — 7ab-\-8x—y-\-z 

Having  set  the  polynomial  which  is  to  be  subtracted  under  the  one 
from  which  the  subtraction  is  to  be  made,  we  conceive  each  term  in 
the  second  line  to  have  its  sign  changed;  then  — a2  added  to  3a2 
makes  2a2  ; — 2 ab  added  to  — 5 ab  makes  — lab  ; and  ox  added  to 
5x  makes  83.  These  results,  together  with  the  — y and  -\-z,  are  con- 
nected in  a polynomial. 

EXERCISES. 

1 From  4a5  + 3c2 — xy,  subtract  2 ab — c2 -\-3xy — 2 z. 

Ans.  2ab-\-4c2 — 4:ry-|-2z. 

2 From  5a2-|-3Z>c  — cd-\-ox,  subtract  2a2 — bc-\-3cd. 

Ans.  3a2+4ic  — 4cd-\-3x. 

3 From  la2b — abc-\- Axy,  subtract  a2b-\-5abc—xy-{-niz. 

Ans.  6a2b  — Gabc-{-5xy — mz 

4 From  8a3 — 4 a2b  — 2Z>c+10,  subtract  3a3+a2&  — o. 

Ans  5a3 — 5a2b—2bcf  15. 

8.  From  3b2c-\-abx-\-2xy2,  subtract  b2c— 3abx+2xy2  — 3. 

.A?zs.  2b~ c 4ubx-\- o . 

6.  From  4a2c  — 3c2 6 + 67/ + 20,  subtract  3c-26+6?/+3iy. 

Ans.  4a2c — 6c26  + 20 — 3 by. 

7.  From  — 3a-\-5b2  —1  xy,  subtract  3b2 -\-2a — xy+bc. 

Ans.  — 5a+2b2  —Gxy—bc. 

8.  From  2ab-{-3bc — 2az2  + 15,  subtract  lab-\-3bc — 10. 

Ans.  — Sab  — 2ax2  + 25. 

9.  From  Sax — 3Z>c+2&2,  subtract  6 ax — be — 3 b2-\-y2. 

Ans.  — ax — 2 bc-\-5b2—y2. 

10.  From  a2b-\-3abc-\- b2c— 7,  subtract  ha2b — abc-\-\b2c 

Ans.  |a26+4«6c+*4b2c—  7 


ADDITION  AND  SUBTRACTION. 


15 


When  one  quantity  is  to  be  subtracted  from  the  Sum  of  two  or 
more  quantities,  the  operation  will  be  expedited  by  changing  the  signs 
of  the  subtractive  quantity,  and  then  adding  all  the  quantities  toge- 
ther. 

11.  From  the  Sum  of  2a+35c — 1,  and  3 a — 4&C+3,  subtract 

4a  — be— 2+i/.  Ans.  a + 4 — y. 

12.  From  the  Sum  of  az2 — &b2  A3  ay2 , and  2aa:2  + 662 — a,y2, 

subtract — 2>ax2 -\-b2  + 2a?/2 — 5.  Ans.  6 ax2 — 62  + 5. 

13.  From  the  Sum  of  2>ab-\-bc — 5x2,  and  ab  — 3 bc-\-%2 — 3,  sub- 
tract 5bc — 3.r2 — 4 Ay2-  Ans.  4 ab — Ibc — £2  + 1 — y2 . 

14.  From  the  Sum  of  5x — 3zy-\-7 y2,  and  5xy — 3 x — 4 y2 — 9,  sub- 
tract —xAxy — 3y2  + 7 . Ans.  3z+£?/+6?/2  — 16. 

15.  From  the  Sum  of  2 a — 5 + 3c<i,  5aA3b  — cd  — 3,  and  a + 25~ 

cd,  subtract  4 a — 5b — cd.  Ans.  4a+95  + 2ecZ — 3. 

16.  From  the  Sum  of  3 x2  Ay — 5xy,  2x2 — 3 yAxy,  and  4 y — 3a;2 

— xy-\- 5„  subtract  x2  -\-5y—xy — 3.  Ans.  x2  — 3y— AxyA3. 

17.  From  the  Sum  of  2aij2  — 2bA3ax,  2ay2-\-b—  ax,  and  3b — 
ay2 A3  — y,  subtract  3ay2  +46— axAy- 

Ans.  — 2Z>  + 3ax  + 3 — 2y. 

18..  From  the  Sum  of  3a2  Axy2  — 2by,  5a2-\-3xy2 — 3 by,  and 
3a;?/2+4a2  + 5y,  subtract  2a2 — xy2 — 56y+5- 

Ans.  10a2  + 8xy2  Aby — 5. 

19.  From  the  Sum  of  a2b2—  3y2-\-5xy,  3a2b2  A3y2 — 2 xy,  and 
5?/2+3a262 — xyA 6,  subtract  a2b2-\-xy — y2 — 3. 

Ans.  6a2£2  + 6?/2+au/+9. 

20.  From  the  Sum  of  5y 2 — 3a*  — 2 be,  4 ax  — 2y2-\-5bc,  and 
3 ax-\-y2 — 5c+5,  subtract  ax— bc-\-3y2  — 1. 

Ans.  i/2  + 3aa;  + 3ic  + 6. 

21.  From  the  Sum  of  a2y — x2-\-bc — d,  ox2 — 3bc-\-5d,  and 
a2y-\-3x2 — 4 be — d,  subtract  ox2 — 2bcA1d. 

Ans.  2a  2 y — 4 be — 4 d. 

22.  From  the  Sum  of  5i2  + 3a*+2y,  3b2 —ax — y-\- 1,  and  4a* 
— b2—5yA 2,  subtract  362+2a* — y+10. 

Ans.  Ab2  +4a*— 3y — 7. 

23.  From  the  Sum  of  7a+6*2 — y2  A 1,  *2+3  y2 — c — 3,  and 
lx2 Ay2 — 3c  + 5,  subtract  5a  + 2x2 — 2_y2  + 5c+2. 

Ans.  2a+12*2  + 5y2 — 9c+l. 

24.  From  the  Sum  of  2 (a — x)Ay,  3 (a — x)  + 2y,  and  5(a — x ) — y. 

subtract  4(a—  x)Ay— 2,  (32).  Ans.  6(a—  x)AyA2- 

25.  From  the  Sum  of  26(a  + c)2+3,  Z>(a+c)2  — 1,  and  3 6(a  + c)2 
— 2,  subtract  the  sum  of  b(a-{-c)2  Ab,  and  5b[aAc)2 — b. 

Ans.  0. 


16 


ADDITION  AND  SUBTRACTION. 


Indicated  Subtraction. 

(37.)  It  is  sometimes  expedient  merely  to  indicate  the  subtraction 
of  a quantity,  without  performing  the  operation. 

To  denote  the  subtraction  of  a positive  monomial,  nothing  more  is 
necessary  than  to  place  the  sign  — before  it;  thus  a — b denotes  that 
b,  that  is,  positive  b,  is  to  be  subtracted  from  a. 

The  subtraction  of  a negative  monomial,  will  he  denoted  by  en- 
closing the  quantity,  with  its  negative  sign,  in  a ( ),  and  prefixing 

the  sign  — to  the  parenthesis. 

Thus  a — ( — b)  denotes  that  negative  b is  to  be  subtracted  from 
a.  When  the  subtraction  is  performed,  the  expression  becomes  a-\-b. 

The  subtraction  of  a polynomial  will  be  denoted  by  enclosing  the 
polynomial  in  a ( ).  and  prefixing  the  sign  — to  the  parenthesis. 

Thus  a — (b-\-c-\-d) ) in  which  the  sign  — denotes  that  the  en- 
closed polynomial  is  to  be  subtracted.  When  the  subtraction  is 
performed,  the  expression  becomes  a — b — c — d. 


Change  of  Signs  in  a Polynomial. 

(38.)  The  signs  of  the  value  of  a Polynomial,  will  evidently  be 
changed,  by  changing  the  signs  of  all  the  terms  of  the  Polynomial. 

The  value  of  a Polynomial  is  therefore  not  affected  by  changing  the 
signs  of  any  or  all  of  its  terms,  enclosing  those  terms  in  a ( ),  and 

prefixing  the  sign  — to  the  parenthesis. 

Thus  a-\-b  — c is  equivalent  to  a — ( — 5 + c),  or  a — (c — b)\  for  if 
(c  — b)  be  subtracted,  as  is  required  by  the  sign  — prefixed  to  it,  the 
result  will  be  a-\-b — c. 

26.  Under  what  different  forms  may  the  value  of  the  polynomial 
ab-\-2cd — 3a; + 5 be  expressed? 

27.  Under  what  different  forms  may  the  value  of  the  polynomia1 
ax  — 3y-f-462  — 5c — 7 be  expressed? 


MULTIPLICATION. 


17 


MULTIPLICATION. 

(39.)  Algebraic  multiplication  consists  in  finding  the  Product  of 
one  quantity  taken  as  many  times,  additively , or  sub tr actively,  as 
there  are  units  in  another  quantity. 

The  quantity  to  he  multiplied  is  called  the  multiplicand,  and  the 
multiplying  quantity  the  multiplier : both  together  are  called  the 
factors  of  the  product,  (12). 

When  the  Multiplier  is  positive,  the  multiplicand  is  repeated 
positively,  or  is  repeatedly  added. 

Thus  5a  X 3,  5a  multiplied  by  positive  3,  is  5a-\-5a-\-5a,  equal  to 
\5a  ; the  multiplicand  5a  being  repeatedly  added. 

When  the  Multiplier  is  negative,  the  multiplicand  is  repeated 
negatively , or  is  repeatedly  subtracted. 

Thus  5a  X — 3,  5a  multiplied  by  negative  3,  is  — 5a  — 5a — 5a, 
which  is  equal  to  — 15a;  the  multiplicand  5a  being  repeatedly  sub- 
tracted 


Multiplication  of  Monomials. 

(40.)  The  Product  of  two  monomials  consists  of  the  product  of  their 
coefficients  multiplied  into  all  their  literal  factors. 

For  example,  3acx2xis  equal  to  6 acx\  for  the  factors  may  be 
taken  in  the  order  3 X 2 acx,  which  becomes  6 acx  by  substituting  G for 
3x2. 

03^  What  is  the  product  of  3a  X 45  ? and  how  do  you  reason  in 
finding  that  product?  What  is  the  Product  of  5a2  b x 2a;  ? and  how 
do  you  reason  in  finding  it?  Of7ac2  Xxy  ? Oixy2x5l  Of  3 X 
la2b2  ? 

(41.)  When  the  same  letter  occurs  in  both  the  monomials  multi- 
plied together,  its  exponent  in  the  Product  will  be  the  sum  of  its  expo- 
nents in  the  two  factors. 

Thus  3a2xX2ax,  or  3a2x1  x2 a1x1,  is  equal  to  6a2axx,  (40).,  and 
this  product  becomes  G a3x2  by  substituting  a3  and  x2  for  their  equiva- 
lents a2a  and  xx,  (13)  and  (16). 

2* 


18 


MULTIPLICATION. 


Observe  that  the  exponents  of  a and  x in  the  product  6 a3x2 , art 
the  sums  of  the  exponents  of  the  same  letters,  respectively,  in  the  two 
monomials  which  are  multiplied  together. 

DP1*  What  is  the  Product  of  4a3  X 2a2  ? Of  3 ax  x 5ax 2 ? Of 
7ax2a3Z>?  Of  4 acxacb'l  Of  a2b3x5bl  Of  8 a2cbxc1.  Of 

a2bxa3b 2 ? 


Sign  of  the  Product. 

(42.)  The  Product  of  two  quantities  is  positive  when  they  both 
have  the  same  sign , and  negative  when  they  have  contrary  signs 
in  other  words,  like  signs  produce  + , and  unlike  signs  produce  — , in 
multiplying. 

When  both  the  quantities  are  positive,  their  product  is  evidently 
positive, — as  in  common  Arithmetic.  Thus  3a  X 2 is  3a+3a,  equal 
to  6a. 

When  both  the  quantities  are  negative,  their  product  is  positive, 
because  such  product  results  from  repeatedly  subtracting  a negative 
quantity , (36). 

For  example,  in  — 3a  X — 2 the  multiplier  — 2 denotes  that  — 3a 
is  to  be  taken  twice  subtractively  ; and  since  the  sign  of  a quantity  is 
changed  in  subtracting  it,  the  product  is  3a  + 3a,  or  6a. 

When  one  of  the  quantities  is  positive  and  the  other  negative, 
their  product  is  negative  because  it  results  from  repeatedly  subtracting 
a positive.,  or  adding  a negative  quantity. 

Thus  in  3a  X — 2,  the  3a  is  to  be  taken  twice  subtractively,  and 
the  product  is  therefore  — 3a — 3a,  equal  to  — 6a. 

[CP1*  What  is  the  Product  of  — 2x  X — 3 ? and  how  do  you  reason 
on  the  sign  of  the  product?  Of  — 3 abx — 2 ? Of  — 5a2  X — 3? 
Of  — x X — 5 ? 

What  is  the  Product,  of  5 axX  — 4 ? and  how  do  you  reason  on  the 
sign  of  the  product  ? Of  4a2cX — 3?  Of  7a:2  X — 5 ? OfZ»2x— 9. 

What  is  the  Product  of  — 2 ah  x3  ? and  how  do  you  reason  on  the 
sign  of  the  product  ? Of  — 3a2at  X 4 ? Of  —3a2  X 2 ? Of  — c3  X 5-1  ? 

What  is  the  Product  of  4x2  X — 7 ? and  how  do  vou  reason  on 
the  sign  of  the  nroduct  ? Of  5ayx  —3?  Of  —3a  X — \o  ? Of  6b 
x— 5? 


MULTIPLICATION. 


19 


When  the  Multiplier  or  the  Multiplicand  is  0. 

(43.)  When  either  of  the  two  factors  multiplied  together  is  0,  the 
product  will  be  0 ; for  it  is  evident  that  0 repeated  any  number  of 
tunes,  produces  only  0 ; and  the  product  is  the  same  when  the  multi- 
plicand and  the  multiplier  are  taken  the  one  for  the  other . 

Thus  5x0  is  0;  axOisO;  («+5  — c)  X 0 is  0. 

The  preceding  principles  and  the  two  following  rules,  embrace  the 
subject  of  Algebraic  Multiplication. 

RULE  III. 

(44. ) To  Multiply  a Monomial  into  a Polynomial. 

Find  the  product  of  the  monomial  into  each  term  of  the  polynomial, 
separately,  and  connect  these  products  in  a polynomial  with  the  proper 
sign  prefixed  to  each,  (40),  (41),  (42). 

EXAMPLE. 

To  multiply  3a5  + 5c2 — 2 xy — 5 by  2 a2x 

3 ab-\-bc2  — 2 xy—p 
2 a2x 


6a3bx-\-2a2bc2x — 4 a2x2y — 10a2a; 

It  will  be  most  convenient  to  multiply  from  left  to  right.  The 
first  term  of  the  multiplicand  is  positive,  and  the  multiplier  being  also 
positive,  the  first  term,  6 a3bx,  of  the  product  is  positive.  In  like  man- 
ner the  second  term,  2 a2bc2,  of  the  product  is  positive,  &c. 

EXERCISES. 

1.  Multiply  2a2  — 35  + e — 2z/  by  3.  Ans.  6a2  — 95  + 3c— 6?/. 

2.  Multiply  3 x2 —xy — y2 4 by  2.  Ans.  6a;2 — 2 xy — 2y2fQ. 

3.  Multiply  4 5 + c2 — 3?/+l  by  — y.  Ans.  — 4 by — c2y-\-3y2 — y 

4 Multiply  —« + 35  — 2a;2 —3  by  —2a;.  Ans.  2 ax — 65a;  + 4a;3  + 6a". 

5.  Multiply  ab2 — 53+  2c — y by  — 5.  Ans.  — 5ab2-\-5b3  — 10c+5y, 

6.  Multiply  — 3+aa: — 1+5;/  by—  a.  Ans.  3a—a2x-\-a—aby. 

7.  Multiply  lx-\-by2  — 4c+d2  by  4.  Ans.  28a;+452/2  — 16c+4i52. 

8 Multiply  a2b—c2  fx— 2 by  3a.  Ans.  3a3b— 3ac2-\-3ax— 6a. 

B* 


20 


MULTIPLICATION. 


RULE  IY. 

(45. ) To  Multiply  a Polynomial  into  a Polynomial. 

Multiply  one  of  the  polynomials  by  each  term , separately,  of  th* 
other  polynomial,  and  add  together  the  several  products. 

EXAMPLE. 

To  multiply  2a2  + 4ac— c2  by  3 a — 5c 

2a2  -f4ac — c2 
3a  —5c 


6a3  -f-  12a2c — 3ac2 

— I0a2c-20ac2  + 5c3 


6a3 -f-  2a2c — 23ac2  + 5c3. 

We  multiply  the  first  polynomial  by  3a,  and  then  by  — 5c,  accord- 
ing to  the  preceding  Rule.  The  two  products  thus  obtained  are  set 
with  similar  terms  one  under  another,  and  added  together. 

The  correctness  of  this  Rule,  as  well  as  of  the  preceding  one,  will 
be  evident  upon  considering,  first,  that  if  each  part  of  a quantity  be 
multiplied,  the  whole  will  be  multiplied ; and,  secondly,  that  if  one 
quantity  be  multiplied  by  each  part  of  another  quantity,  the  former 
will  be  multiplied  by  the  whole  of  the  latter. 


EXERCISES. 


9.  Multiply  a2  — 2a  + 5 by  a-f-3. 

10.  Multiply  2x2-|-3x — 1 by  x — 5. 

11.  Multiply  a2  — 2 az-j-.i2  by  a + x. 

12.  Multiply  x2  -\-3xy-\-y2  by  x — y. 

13.  Multiply  h2 — 3 be — c2  by  2 b-\-c. 

14.  Multiply  2a2  + 3ac — c2  by  a — 2c. 

15  Multiply  2 b2 — 2 bx-\-x2  by  2b — x. 

16  Multiply  b3  -\-b2-\-b  by  b3-\-b2. 

17  Multiply  x2  — 2c2  -\-2y  by  c2  -\-y. 

15.  Multiply  a2  -f  bxpy  by  a2  — bx. 


Am.  a3 -fa2  — a+15 
Ans.  2z3  — lx2  — 16x-(-5. 
Ans.  a3 — a2x — ax2-\-x3 
Ans.  x3  + 2 x2y — 2xy2—y3 
Ans.  2 b3 — 5b2c — 5bc2—c3 
Ans.  2a3  — a2c— lac2  + 2c3 
A?is.  4 b3 — &b2 xd-4bx2  — x:' 
Ans.  b6 -\-2b5 -j-2b* Ab3. 
M/iS.  c2a'2 — 2c4  -\- x2 y-\-2y2 
Ans.  a*-\-a2y — b2,i 2 — bi y. 


DIVISION. 


21 


I 3 Find  the  Product  of  (2a2  — 4a5-|-262)(3a  — 2b). 

Ans.  6a3  — I6a26+14a62 — 463. 

20  Find  the  Product  of  (3a;2 — 2au/+2z/2)(2a;2  -\-3,xy). 

Ans.  6a :i-\-5x3y — 2x2y2  -\-3xy3 . 

21  Find  the  Product  of  (2a2+3a5 — b2){a 2 — ad  + 62). 

Ans.  2ai-\-a3b — 2a262+4a63 — 6*. 

22  Find  the  Product  of  (3a;2 — 2xy+5)(x2  + 2xy— 3) 

A7is.  3a;4 +4 x3y — 4a;2  — 4a;2 y2  + 1 Gxy — 15. 
23.  Find  the  Product  of  the  three  polynomials  (a+6),  (a— b),  and 
(a1  \-ab-\-b2)  Ans.  a4+a36 — ab3—b^. 


DIVISION- 

(46.)  Algebraic  division  consists  in  finding  a factor  or  Quotient, 
which,  multiplied  into  a given  divisor,  will  produce  a given  dividend. 

Thus  6a2z  — 2a,  6a2 x divided  by  2a,  gives  the  quotient  3 ax,  be- 
cause this  factor,  multiplied  into  the  divisor  2a,  produces  the  dividend 
6 a2x. 

Division  of  Monomials. 

(47. ) The  Quotient  of  two  monomials  will  be  found  by  dividing 
the  coefficient  of  the  divisor  into  that  of  the  dividend,  and  subtracting 
the  exponents  in  the  divisor  from  those  of  the  same  letters,  respectively,, 
in  the  dividend. 

Thus  \3a2x2y-^5oJ2x  gives  the  quotient  2axy,  because  this  quo- 
tient, multiplied  into  the  divisor  5a2x,  produces  the  dividend  1 0a3z2y ; 
and  2 axy  is  found  by  dividing  5 into  10,  and  diminishing  the  expo- 
nents of  a and  x in  the  dividend  by  the  exponents  of  a and  x in  the 
divisor. 

The  dividend  being  the  product  of  the  divisor  and  quotient,  the 
exponents  in  the  dividend  are.  the  sums  of  the  exponents  of  the  same 
letters  in  the  divisor  and  quotient,  (41);  hence  the  exponents  in  the 
quotient  will  always  be  found  by  subtracting  as  above. 

(UP3  What  is  the  Quotient  of  6a3x2y—  3ax  1 and  how  do  you 
prove  that  quotient  true  ? What  is  the  Quotient  of  4 ab2c-t-bt  and 
how  do  you  prove  that  quotient  true  ? What  is  the  Quotient  of 
ia2hz  — 2a  '?  and  how  do  you  prove  that  quotient  true  ? 


22 


DIVISION. 


When  any  Exponent  is  reduced  to  0. 

(48.)  Any  quantity  whatever  with  exponent  0 is  equivalent  to 
unity ; and  a factor  with  this  exponent  may  therefore  he  canceled. 

For  example,  a2 —a2  gives  the  quotient  a0,  by  subtracting  the  ex- 
ponent of  the  divisor  from  that  of  the  dividend. 

But  any  quantity  contains  itself  once,  and  therefore  a2  —a2  also 
gives  the  quotient  1 ; and  these  two  quotients,  a0  and  1,  must  be  equi- 
valent, to  each  other. 

Since  a in  the  preceding  illustration  may  represent  any  quantity  we 
please,  any  quantity  whatever  with  exponent  0,  is  equivalent  to  1, 
or  is  a symbol  of  unity. 

To  divide  10 a3b  by  5a3. 

Subtracting  the  exponent  in  the  divisor  from  the  exponent  of  a in 
the  dividend,  we  find  the  quotient  2 a°b  The  factor  a0  is  equivalent 

to  1,  and  may  therefore  be  canceled  without  affecting  the  value  of  the 
quotient. 

The  quotient  of  10 a3b  — 5a3  may  therefore  be  given  under  the  two 
different  forms  2 a°b  and  2b. 

[FP*  Under  what  two  different  forms  may  the  Quotient  of  6a2 z 
2a2 be  represented  ? Of  8«62  — 2b2  1 Of  10 abc  — 5al  Of  9 ax2  — 
x2  ? Of  7 abc2  -h7  c2  ? Of  5ax-^axl  Of  3 aby2 —ay2  l Of  10a:2- 
x2  ? 

Sign  of  the  Quotient. 

(49.)  The  Quotient  of  two  quantities  is  positive  when  they  both 
have  the  same  sign,  and  negative  when  they  have  contrary  signs  ; — 
in  other  words,  like  signs  produce  and  unlike  signs  produce  — , in 
dividing. 

Thus  -\-ax^-  -\-a  gives  +:r,  because  +aX  + x is  -j-  ax; 

— ax-. a gives  +x,  because  — a X is  — ax  ; 

-\-ax-. a gives  — x,  because  — aX  — x is  +ax; 

and  — ax~-\-a  gives  — x,  because  +«x  — # is  — ax  ; (42.) 

From  these  examples,  it  will  be  seen  that  the  sign  of  the  quotient 
must  be  such  that,  the  quotient  multiplied  into  the  divisor  shall  pro-  ■ 
duce  the  dividend. 

It.  thus  appears  that  the  rule  for  the  sign  of  the  quotient,  is  the 
same  as  that  for  the  sign  of  the  product,  of  two  quantities.  The  ■ 
learner  must  be  careful  not  to  apply  this  rule  in  finding  the  sum,  or 
difference,  of  two  quantities. 


DIVISION. 


23 


03=*  What  is  the  Quotient  of  4<?2;r-4-2?  and  how  do  you  reason 
on  the  sign  of  the  quotient?  Of  6ac2  — 2c  1 Of  9x3y^‘3xy  ? Of 
a3  H-a  ? 

What  is  the  Quotient  of  — la3-. a2  ? and  how  do  you  reason 

on  the  sign  of  the  quotient  ? Of  — 12o2  4 ? Of  — a3b~. ab  ? Of 

-b^b21 

What  is  the  Quotient  of  8a2-: a 2 ? and  how  do  you  reason  on 

the  sign  of  the  quotient?  Of  10 ab-. — -2b  1 Of  4a2c-: al  01 

O.r2  ~ -x2  ? 

What  is  the  Quotient  of  — 7a3-f-7a  ? and  how  do  you  reason  on 

the  sign  of  the  quotient?  Of  — 12ac-t-3a?  Of  — 9a36-^-3?  Of 

— 20«4^4? 

What  is  the  Quotient  of  10«3  -^5«?  and  why  is  that  quotient 

true  ? What  is  the  Quotient  of  L2a:2  3 ? and  why  is  that  quotient 

true  ? What  is  the  Quotient  of  — 9a3by  — 3a3  ? and  why  is  that 

quotient  true  ? What  is  the  Quotient  of  — 20a4a:3  5 ax  1 and  why 

is  that  quotient  true  ? What  is  the  Quotient  of — lOOax y-. xy  ? 

and  why  is  that  quotient  true  ? 

When  the  Dividend  or  the  Divisor  is  0. 

(50.)  The  quotient  of  0 divided  by  any  quantity,  is  0 ; but  the 
quotient  of  any  quantity  divided  by  0,  is  infinitely  great. 

First , let  a represent  any  quantity  we  please  ; then 
0—  a gives  the  quotient  0,  because  this  quotient  multiplied  into  the 
divisor  a , produces  the  dividend  0,  (43). 

Secondly,  a divisor  may  be  taken  so  small  as  to  be  contained  a 
great  number  of  times  in  any  given  dividend,  represented  by  a ; if  the 
divisor  be  still  diminished,  the  quotient  will  be  increased  ; and  if  the 
divisor  be  diminished  ivithout  limit , the  quotient  will  be  increased 
without  limit. 

If  therefore  the  divisor  were  dimished  to  0,  the  quotient  would  be 
unlimited,  that  is,  infinitely  great. 

The  character  oo  is  the  sign  of  infinity ; 

we  have  then  0 -^-a  equal  to  0 ; and  <z-^-0  equal  to  go,  infinity 

The  preceding  principles,  and  the  two  following  Rules,  embrace 
the  subject  of  Algebraic  Division,  when  the  Quotient  is  an  integral 
quantity 


24 


DIVISION. 


RULE  V. 

(51.)  To  Divide  a Monomial  into  a Polynomial. 

Find  the  quotient  of  the  monomial  divided  into  each  term  of  tha 
polynomial,  separately,  and  connect  these  quotients  in  a polynomial, 
with  the  proper  sign  prefixed  to  each.  (47),  (48),  (49). 


EXAMPLE. 

To  divide  20a2a:  — 15aa;2+30aa:y2 — 5ax  by  5ax. 

5ax  J20a2x — 15aa:2  +30  axy2  — 5ax 
4 a — 3a:  + 6 y2  —1 


Dividing  into  the  first  term  of  the  dividend,  we  find  the  quotient 
term 'la]  dividing  into  the  second  term  of  the  dividend,  we  find  the 
quotient  — 3a:;  and  so  on,  through  the  dividend. 


EXERCISE  S. 


1.  Divide 

2.  Divide 

3.  Divide 

4.  Divide 

5.  Divide 

6.  Divide 

7.  Divide 

8.  Divide 

9.  Divide 
10.  Divide 
1 1 Divide 

12.  Divide 

13.  Divide 

14.  Divide 

15.  Divide 


3a:3-f-6x2-f3aa; — 15a:  by  3a:.  Ans.  x2-\-2z-\-a — 5. 

2 ab  — 6a2a?-{-8a3y — 2a  by  2a.  Ans.  b — 3aa;-f-4a2y — 1 
14a2 — lab  \-2\ax — 21a  by  7a.  Ans.  2 a — 5 + 3a:  — 3. 
a3x+3a2a;2 — 6ax3+3axbv  a.  Ans.  a2x+3ax2 — 6x3  + 3x. 
552  — 1053 + 554y — 155s  by  5b.  Ans.  b—2b2+b3y — 354. 
bx2 -\-2x3 — 8cx4  +7xs  by  x2 . Ans.  b-\-2x  — Sca-2  + 7a-3. 
4a4 — 8a3 — 4a25  + 8a  by  2a.  Ans.  2 a3 — 4a2—  2a5-f  4 

ay4  + a2y3 — a3y 2 — ay  by  ay.  Ans.  y3+ay2 — a2y — 1. 

Ans.  1 — 5y+5y2— 3y3 
Ans.  b3  — 352+45— 5 
alas,  c3 — 3c2+6c— 1 
alas.  Sy2 — 4a+a2y — 3 
alas.  2— 4a+3a2— 4. 
alas,  a2 — ab-\-a2b— a3. 


- a3y 2 — ay  by  ay. 
— y+5y2— 5y3+3y4  by  — y. 
353  — 952  + 125— 15  by  + 3. 

— c5 +3c4  — 6c3  + c2  by  — c2. 
8y3  — 4ay  + a2y2  — 3y  by  y. 

— 10  + 20a-15a2  + 20  by  —5. 
a3b  — a252  + a352 — a45  by  ab. 


4y4+a:3y3 — x2y2  — xy by a?y.  alas,  x3 ,?3  —x2  2 —r y— ' 


DIVISION. 


25 


RULE  VI. 

(52.)  To  Divide  a Polynomial  into  a Polynomial. 

1 . Arrange  the  divisor  and  dividend  according  to  the  ascending  or 
the  descending  poxcers  of  the  same  letter,  (22). 

2.  Divide  the  first  term  of  the  divisor  into  the  first  term  of  the 
dividend  ; multiply  the  whole  divisor  by  the  quotient  term  ; subtract 
the  product  from  the  dividend ; divide  into  the  remainder,  as  before, 
and  so  on, — observing  to  connect  the  several  quotient  terms  in  a poly- 
nomial, with  the  proper  sign  prefixed  to  each. 

EXAMPLE. 

To  divide  6a4*d-3a3*K — 4a2*4 +*6  by  2ax-\-x 2 

2 a*+*2  ) 6a4*2  + 3a3*3 — 4a2*4-f-*6  ( 3 a3x — 2a*3+*4 
6a4*2  + 3a3*3  . 


— 4 a2*4-|-*6 
— 4a2*4  — 2a*5 


2 a*s+*6 
2ax5  -j-x6 


The  divisor  and  dividend  are  here  arranged  according  to  the  de- 
scending powers  of  a,  or  the  ascending  powers  of  x. 

The  first  term  2a*  of  the  divisor,  divided  into  the  first  term  6a4*2 
if  the  dividend,  gives  the  quotient  term  3a3*.  Multiplying  the  whole 
divisor  by  this  quotient  term,  and  subtracting  the  product  from  the 
dividend,  the  remainder  is  — 4a2*4 4-*6. 

We  next  divide  the  first  term  of  the  divisor  into  the  first  term, 
— 4a2*4,  of  the  remainder,  and  find  the  quotient  term  — 2a*3  ; k c. 

By  this  Rule  the  divisor  is  multiplied  by  each  part  of  the  quotient. 
and  the  successive  products  subtracted  from  the  dividend.  When  the 
dividend  is  thus  exhausted,  the  product  of  the  divisor  and  quotient  is 
equal  to  the  dividend  ; and  the  quotient  is  thus  proved  to  be  correct. 

The  Divisor,  when  a polynomial,  is  sometimes  set  on  the  left  of  the 
dividend,  and  the  quotient  under  the  divisor.  This  arrangement  is 
thought  to  afibrd  greater  facility  in  multiplying  the  divisor  by  the 
quotient  term. 


26 


DIYISIOH. 


EXERCISES. 


16.  Divide  a2  — 2aa-|-a2  by  a— a. 

17.  Divide  a:3 — 3aa2+3 a2x — a3  by  a:— a. 

18.  Divide  6a4  + 9a2 — 15a  by  3a2 — 3a. 

19.  Divide  a4 +a2a2 -|-a4  by  a2—  ax-\-x2. 

20.  Divide  2a3  — 19a;2  -|- 26a — 16  by  a:  — 8. 

21.  Divide  4a;4 — 5a2c2+c4  by  2a"2  — Sae+c2. 

22.  Divide  a4— 2a2a2+a4  by  a2  + 2aa+£2- 

23.  Divide  12 — 4a  — 3a2+«3  by  4 — a2. 

24.  Divide  4y4  — 9y2-{-6y — 1 by  2y2 — 1. 

25.  Divide  2a;4 — 32  by  x — 2. 

26.  Divide  4a5  — 64a  by  2a  — 4. 


Ans.  a — x. 
Ans.  x2 — 2 aa+a2. 
Ans.  2a2 -f- 2a + 5 
Ans.  a2+a a + a2. 
Ans.  2x 2 — 3a; -(-  2. 
Ans.  2a2  +3arc+c2. 
Ans.  a 2 — 2 ax-f-x2. 
Ans.  3 — a. 
Ans.  2y2  — 3?/+l. 
Ans.  2a3 +4a2  + 8a+ 16. 
Ans.  2a4  + 4a3  + 8a2  + 16a. 


27.  Divide  6 y 6 — 96 y2  by  3 y — 6.  Ans.  2//5+4//4  + S//3-f-16//2. 

28.  Divide  a4 + 4a4  by  a2 — 2aa-f2x2.-  Ans.  a2 2aa+2a2. 

29.  Divide  a5 — x5  by  a4 -\- a2 x-\-a2x2 -\-axz -\-x4 . Ans.  a — x. 

30.  Divide?/4 +4?/2s2 — 32s4  by y-\-2z.  Ans.y 3 — 2y2z-\-Qyz2 — 16s3. 


(53.)  The  indicated  Product  of  two  or  more  factors  is  divided  by 
any  quantity,  when  either  of  the  factors  is  divided  by  that  quantity. 

Thus  to  divide  3a2a(a2+?/2)  by  3a,  we  divide  the  factor  3a2a,  | 
and  find  the  quotient  a 2 (a2  + ?/2). 

In  the  following  exercises,  the  first  of  the  given  factors  may  be  1 
divided  by  the  given  divisor. 

31.  Divide  (a2 -\-2ay-\-y2)(b2 — cd2  +3)  by  a+ y. 

Ans.  (a  + ?/)(£3 — c<22+3).  | 

32.  Divide  (2a3— 6aa2  + 6a2a — 2a3)  (c2-f-3cy — y2)  by  a — a. 

alas.  (2a2 — 4aa+2a2)(c;2  + 3c7/ — y2). 

33.  Divide  (a4  + 4?/4)  (3a?/2—  5//3+3?/4  — 4)  by  a2  -2ay-\-2y2 . 

Ans.  (a2  -j-2a?/-f-  2y2)  (3 xy2 — 5?/3  + 3?/4 — 4.) 

34.  Divide  (8a4 — 2a3a — 13a2a2 — 3aa3)  (y2-\-2xy)  (^3  + 3a2  2 — a3| 
by  4a2~f-5aa-j-a2.  .A?£S.  (2a2 — 3ax)  (?/2-{-2a'?/)  (v/3-f-3a2//2 — a3) 


27 


CHAPTER  III. 

COMPOSITE  QUANTITIES.— COMMON  MEASURE.— COMMON  MULTIPLE. 

Composite  and  Prime  Quantities. 

(54.)  A composite  quantity  is  one  which  is  the  product  of  two  fac- 
tors, each  differing  from  unity ; and  a.  prime  quantity  is  one  which  is 
not  the  product  of  such  factors. 

Thus  3a2  is  a composite , while  a-\-b  is  a.  prime  quantity. 

Is  aib  a composite  or  a prime  quantity  1 Is  a+5  a composite 
or  a prime  quantity  ? a2  1 7 x2  1 2a-\-'3b'l  5a — 5c2  1 

Decomposition  of  Quantities. 

(55.)  Decomposing  a composite  quantity  consists  in  resolving  it 
into  its  factors.  Any  divisor  of  the  quantity,  and  the  corresponding 
quotient,  are  two  factors  into  which  it  may  be  resolved,  (46.) 

Thus  if  the  binomial  3x-{-€>ax  be  divided  by  3a;,  the  quotient  will 
he  l + 2a  ; the  binopiial  may  therefore  be  resolved  into  the  factors 

3a;(l  + 2«),  3a;  into  the  binomial  l + 2a. 

Resolve  2«  + 4aa;  — 6a2a;2  into  component  factors. 

Resolve  a2x — 3ax2  -\-8ay2  into  component  factors. 

Resolve  4 a2-\-a2b  — 5a2 y into  component  factors. 

Resolve  2a3 — 3ux-\-l  a*y  into  component  factors. 

Resolve  5ax-\-5a2 x — I0a3a3  into  component  factors. 

Resolve  lab  — I4ab2  — liabx  into  component  factors. 

A monomial  factor  may  generally  be  found  by  mere  inspection,  and 
a composite  polynomial  be  resolved  into  a monomial  and  a polynomial 
factor,  as  above. 

No  general  Rule  can  be  given  for  resolving  a polynomial  into  the 
polynomi  d factors  of  which  it  may  be  composed.  But  there  are  par- 
ticular Binomials  which  have  well  known  binomial  divisors , by  means 
of  which  such  Binomials  may  be  resolved  into  two  factors,  (55). 

The  following  divisions  will  be  found,  on  trial,  to  terminate  with- 
out remainders. 


28 


COMPOSITE  QUANTITIES. 


(56.)  The  Difference  of  two  quantities  will  divide  the  difference  of 
any  powers  of  the  same  degree  of  those  quantities. 

Thus  a — b will  divide  a 2 — b 2,  or  a3 — b3,  or  a4 — b 4,  &c. 

(57.)  The  Sum  of  two  quantities  will  divide  the  sum  of  any  odd 
'powers  or  the  difference  of  any  even  powers,  of  the  same  degree,  oi 
those  quantities. 

Thus  a-\-b  will  divide  o3+d3,  or  a5  +55 , or  a7  -\-b7 , &c. , 
also  a-\-b  will  divide  a 2 — b2,  or  a 4 — d4,  or  a6 — b6,  See.; 

The  factors  of  certain  Binomials  and  Trinomials  may  also  be 
known  from  the  manner  in  which  the  products  and  squares  of  binomials 
are  formed.  This  will  be  seen  in  the  following  propositions. 

(58.)  The  Product  of  the  sum  and  difference  of  two  quantities  is 
equal  to  the  difference  of  the  squares  of  those  quantities. 

Thus  (a+i)  ( a — b)  is  equal  to  a 2 — b2  ; and  this  last  binomial  may 
therefore  be  divided  by  either  a-\-b  or  a — b. 

This  proposition,  it  will  be  observed,  is  included  in  the  two  pre- 
ceding ones,  (56.)  (57.) 

ttsP*  What  is  the  Product  of  (a+x)(a — x)  ? What  is  the  Pro- 
duct of  (<z+5)  (a  — 5)  ? Of  (3+?/)  (3— i/)  ? Of  (x — l)(x+l)? 

(59.)  The  Square  of  the  sum  of  two  quantities  is  equal  to  the  sum 
of  the  squares  plus  twice  the  product  of  the  two  quantities. 

Thus  (af-b)  (a+6),  that  is,  the  square  of  a-\-b,  is  equal  to 
a2  -\-b2  -\-2ab  or  a2  -f2ab-\-b2 . 

This  trinomial  may  therefore  he  divided  by  a-\-b. 

What  is  the  Product  of  (a  + x)  (a  + x),  or  the  square  ofa-f  xl 
What  is  the  Square  of  a -\-y  ? Ofx  + 3?  Of  «+c  ? Ofy+ol 

(GO.)  The  Square  of  the  difference  of  two  quantities  is  equal  to  tb> 
sum  of  the  squares  minus  twice  the  product  of  the  two  quantities. 

Thus  (a — b)  ( a — d),  that  is,  the  square  of  a — b , is  equal  to 
a2  -\-b2 — 2 ab  or  a2 — 2ab-\-b2. 

This  trinomial  may  therefore  be  divided  by  a — h. 

[CP*  What  is  the  Product  of  ( a — x)(a — x).  or  the  square  of 
a — xl  What  is  the  Square  of  a — yl  Ofy — 2?  Ofd— x?  Of  1 — y! 


COMPOSITE  QUANTITIES. 


29 


Tlie  preceding  principles  will  be  found  applicable  to  the  following 
EXERCISES. 

1.  Resolve  a 2 — x2  into  component  factors. 

Ans.  ( a — x ) (a  + a:) 

2.  Resolve  a3  +?/3  into  component  factors. 

Ans.  (a-{-y)(a2 — ay-\-y2) 

3.  Resolve  a 3 — a;3  into  component  factors. 

Ans.  ( a — x)(a2 -\-ax-\-x2) 

4.  Resolve  a4 — yi  into  component  factors. 

A?is.  (a2-\-y2)  (a-\-y)  (a — y). 

5.  Resolve  a3 — 8a:3  into  component  factors. 

ai«s.  (a — 2x)(a2 -\-2ax-\-4x2). 

6.  Resolve  i —&y3  into  component  factors. 

Ans.  (1  — 2y)  (l  + 2y+4?/2) 

7.  Resolve  l+27a:3  into  component  factors. 

Ans.  (l+3a:)(l  — 3a:+9a:2). 
8 Resolve  8a3 — 27 y3  into  component  factors. 

Ans.  (2a — 3y)  (4a2 -\-6ay+9y2). 

9.  Resolve  a3x3  ~\-c3y3  into  component  factors. 

A?is.  (ax-\-cy)(a2x2 -axcy-\-c2y2). 

10.  Resolve  a4  — 16a:4  into  component  factors. 

Ans.  (a2  -f-4a:2)  (a-\-2x)  (a  — 2a:). 

11.  Resolve  a2 -|-2aa:-f-a:2  into  component  factors. 

Ans.  (a  + a:)  (a  + a:). 

12.  Resolve  a2  — 2ay-\-y2  into  component  factors. 

Ans.  (a—y)(a—y). 

13.  Resolve  a2 -\-4ax-i-4x2  into  component  factors. 

Ans.  ( a-\-2x ) (a+ 2a:). 

14.  Resolve  9a2 — 6 ay-\-y2  into  component  factors. 

A?is.  (3a— y)  (3a — y). 

15.  Resolve  4a2  + 12ax-\-9x2  into  component  factors. 

Ans.  (2a  + 3a;)  (2a+3x) 

16.  Resolve  a2a;2 — 2ax-\-  \ into  component  factors. 

Ans.  (ax — 1)  (aa:—  1) 

17.  Resolve  \-\-4xy-\-4x2y 2 into  component  factors. 

Ans.  ( 1 + 2a:y)  ( 1 + 2xy) 

18.  Resolve  4a2 x2  — \2abxy-\-9b2  y2  into  component  factors. 

Ans.  (2 ax  — 3by) (2ax — 3 by) 

19.  Resolve  9aixi  -\-24a2cx2y-{-  16c2?/2  into  component  factors. 

Ans.  (3a2x2  + 4q/)  (3a2 x2  +4 cy). 

20.  Resolve  16a432 — 32a2 czy2  -f- 1 6c2?/4  into  component  factors. 

Ans.  (4 a2x — 4cy2)(4a2x—4cy2) 


30 


COMPOSITE  QUANTITIES. 


(61.)  A Trinomial  may  be  resolved  into  two  binomial  factors, 
whenever  one  of  its  three  terms  is  a square , another  the  sum  of  the 
products  of  the  square  root  of  that  term  multiplied  into  any  two  quan- 
tities, and  the  remaining  term  the  product  of  the  same  two  quantities 

For  example,  let  it  be  required  to  decompose  the  trinomial 
a2  — a — 12. 

The  first  term  is  the  square  of  a ; the  second  term  is  the  sum  oj 
the  products  of  a multiplied  into  3 and  — 4 ; and  the  third  term  is  the 
product  of  3 and  — 4. 

Now,  from  the  manner  in  which  the  product  of  two  binomials  is 
formed,  it  is  evident  that 

a2  — a— 12  is  equal  to  (<z  + 3)  ( a — 4). 

In  like  manner  the  following  Trinomials  may  be  decomposed. 

21.  Resolve  «2  + 7a+12  into  component  factors. 

Ans.  (a+3)(a+4). 

22.  Resolve  x2-\-x  — 30  into  component  factors. 

Ans.  (or-J-  6)  (ar — 5).  , 

23.  Resolve  a 2 — 8 a — 20  into  component  factors. 

Ans.  (n  — 10)(a+2).  | 

24.  Resolve  y 2 — 1 0 y/  — 39  into  component  factors. 

Ans.  (y— 13).(y+3). 

25.  Resolve  a2x2-j- 9<zir+18  into  component  factors. 

+«.s.  (<zar+6)  (ax+3) 

26.  Resolve  a2 y2  — llay  + 28  into  component  factors. 

Ans.  (ay— 7)  (ay— 4) 

27.  Resolve  3ab2  + 21a6+36a  into  component  factors. 

The  given  quantity  may  be  resolved  first  into 

3a(62+76+12); 

ami  the  trinomial  factor  thus  obtained  may  be  resolved  into  (5  + 3) 

(6  + 4) 

3a  (6+ 3)  (6+ 4).  I 

28.  Resolve  5x2—5x  — 60  into  component  factors. 

Ans.  5 (x— 4)  (:t+3).  I 

29.  Resolve  2«62  + 18a6  + 36a  into  component  factors. 

Ans.  2a  (6+6)  (6+3).  j 

30.  Resolve  4ax3  + 10aa:2  + 6a.'c  into  component  factors. 

Ans.  aa:(2x+3)  (2.r+  2).  I 


COMMON  MEASURE. 


31 


COMMON  MEASURE. 

(G2.)  One  quantity  is  called  a measure  of  another,  or  is  said  to 
measure  another,  if  it  will  divide  the  other,  without  a remainder ; 
and 

A common  measure  of  two  or  more  quantities  is  any  quantity  that 
will  divide  each  of  them,  without  a remainder. 

Thus  5a  is  a common  measure  of  10a6  and  5a2+  1 5ab2 . 

[CP*  Name  a Common  measure  of  4 a2b  and  6 ab2 . Of  9a2  and 
3 a2+a2c.  Of  5abc-\-3a2c  and  4ac.  Of  a2x  and  a2x-\-a3y. 

, Greatest  Common  Measure. 

(G3.)  The  greatest  common  measure  of  two  or  more  quantities,  is 
the  greatest  quantity  that  will  divide  each  of  them,  without  a 
remainder. 

Thus  3a2  is  the  greatest  common  measure  of  6a2 xfoa3 y and 
Za2b — 6 a3cd,  since  it  is  the  greatest  quantity  that  will  divide  each  of 
these  binomials,  without  a remainder. 

Here  it’  may  he  remarked,  that  when  we  speak  of  algebraic  quan- 
tities as  being  greater  or  less,  we  have  reference  to  the  coefficients  and 
exponents  of  the  same  letters,  and  not  to  any  particular  values  which 
the  letters  may  be  supposed  to  represent. 

Thus  a2  is,  algebraically,  always  greater  than  a ; though  its  nu- 
merical value  would  be  less  than  that  of  a,  if  the  latter  should  be 
taken  to  represent  a.  proper  fraction,  as  •£,  f,  &c. 

dp"  What  is  the  Greatest  Common  Measure  of  ‘2>a2x-{-ia'3x  and 
d3y — 5a2 y3  1 Of  5a3c+  10a3c2  and  10a2b — 5a3bxl  Of  2 ax2  — 

6 a2y  and  4a2a:+8a?/2 — 2a  l 

When  two  quantities  have  no  common  measure  but  unity , they 
are  said  to  be  prime  to  each  other. 

(64.)  The  greatest  common  measure  of  two  or  more  quantities  is 
composed  of  all  the  factors  which  are  common  to  those  quantities , that 
is,  all  the  factors  which  are  found  in  each  quantity. 

For  example,  the  common  factors  in  ?>axy2  and  6a2y2z  are  3,  a, 
and  y2  ; and  3 ay2  is  the  greatest  common  measure  of  the  two  quan 
tities. 


COMMON  MEASURE. 


R 2 

But  as  no  general  Rule  can  be  given  for  resolving  a Polynomial 
into  the  'polynomial  factors  of  which  it  may  be  composed,  we  make 
the  Rule  for  finding  the  greatest  common  measure  depeud  on  the  fol- 
lowing proposition  ; — 

(65.)  The  greatest  common  measure  of  two  or  more  quantities,  is 
the  same  as  that  of  the  least  of  those  quantities  and  the  remainder,  or 
remainders,  if  any,  after  dividing  the  least  into  the  other,  or  each  of 
the  others. 

We  may  suppose  any  two  quantities  to  be  represented  by 
a and  na-\-b  ; 

n being  some  integral  number,  and  b less  than  a. 

It  is  piain  that  any  measure  of  a will  also  measure  na , n times  a ; ) 
and,  measuring  na,  if  it  measure  ?ia-\-b,  it  must  also  measure  b. 
Hence  there  can  be  no  common  measure  of  a and  na-\-b  which  is  not  < 
a common  measure  of  a and  b ; the  greatest  common  measure,  there-  | 
fore,  of  a and  b,  will  be  the  greatest  common  measure  of  a and  { 
na-\-b. 

But  b is  the  remainder  after  dividing  a into  na+b ; hence  the 
proposition  is  true  for  two  quantities ; and  in  like  manner  it  may  tx»  ! 
proved  for  three  or  more  quantities. 

RULE  VII. 

(GG.)  To  find  the  Greatest  Commo?i  Measure  of  two  Quantities.  ] 

1.  Divide  one  of  the  quantities  into  the  other,  and  the  remainder  i 
into  the  divisor,  and  so  on,  until  there  is  no  remainder  : the  last  _ risor  i 
will  be  the  greatest  common  measure  required. — But, 

2.  When  any  factor  is  contained  in  each  term  of  the  divisor,  with-  i 
out  being  contained  in  each  term  of  the  corresponding  dividend, — 
such  factor  must  be  canceled,  before  dividing. — And 

3.  When  the  first  term  of  the  divisor  is  not  a measure  of  the  first  I 
term  of  the  dividend — multiply  the  several  terms  of  the  latter  by  some  ij 
quantity  which  will  thus  render  its  first  term  divisible,  without  a 
remainder. 

Note — Tli;s  Rule  might  be  readily  extended  to  three  or  more  quantities;  I 
but  we  seldom  or  never  have  occasion  to  find  the  greatest  common  measure 
of  more  than  two  algebraic  quantities. 


COMMON  MEASURE. 


33 


EXAMPLE. 

To  find  the  greatest  common  measure  of  the  polynomials 
4a2x — 5ax2-\- x3  and  3a3 — 3 a2x-\-ax2 — x3. 

4 a2x — 5ax2-\-  x3  3a3 — Sa2x-\-  ax2 — x 3 


4a2  — 5ax  -\-x2 


4 


12a3  — 12a2a:+  4 ax2 — 4a;3  ( 3a 
12a3  — 15a2a:+  3 ax2 


19aas2  — 19x3 


a — x 


3a2x+  ax2 — 4a:3 
4 


12a2a:+  4aa:2 — 16a:3  ( 3a; 
12a2a: — 15aa:2-f-  3a:3 


19aa:2  — 19a:3 


4a2  — 5ax-\-x2  ( 4a — x 
4 a2  — 4aa: 


— ax-\-x2 
— ax-\-x 2 


In  this  example  the  factor  x is  contained  in  each  term  of  the 
divisor  4 a2x — 5ax2  -\-x3,  without  being  contained  in  each  term  of  the 
dividend  ; we  therefore  cancel  this  factor,  and  take  4a2 — 5ax-\-x2  for 
a divisor. 

Then,  the  first  term  4a2  of  this  divisor  not  being  a measure  of  the 
first  term  3a3  of  the  dividend,  we  multiply  the  dividend  by  4,  which 
renders  the  first  term  12a3  divisible,  without  a remainder. 

We  also  multiply  the  remainder  3 a2x-\-ax2 — 4a:3,  still  regarded 
as  a dividend,  by  4,  to  make  the  first  term  divisible,  without  a remain- 
der. 

In  the  next  remainder,  19aa:2  — 19a:3,  the  exponent  of  a,  the  letter 
of  arrangement , being  less  than  in  the  first  term  of  the  divisor, — we 
divide  this  remainder,  after  canceling  the  factor  19a:2,  into  the  formei 
divisor,  which  now  becomes  the  dividend. 

The  divisor  a — x completes  the  operation,  and  is  the  greatest  com- 
mon  measure  required. 


3 


34 


' COMMON  MEASURE. 


It  remains  to  elucidate  the  Rule. 

The  -greatest  common  measure  of  the  two  polynomials,  is  the  same 
as  that  of  the  divisor  4 a2x — 5ax2-\-x 3 and  the  remainder  after  the 
first  division,  (65).  On  the  same  principle  the  greatest  common  mea- 
sure of  the  first  divisor  and  remainder,  is  the  same  as  that  of  the  first 
remainder  and  the  second  remainder  ; and  so  on,  to  the  last  remain- 
der, which  becomes  the  last  divisor. 

Hence,  the  last  remainder,  or  divisor — being  the  greatest  common 
measure  of  itself  and  the  preceding  divisor,  and  so  on  to  the  first  re- 
mainder and  divisor — is  the  greatest  common  measure  of  the  given 
polynomials. 

Again  ; since  the  greatest  common  measure  is  composed  of  all  the 
factors  found  in  each  of  the  two  polynomials,  (64),  it  is  not  affected  by 
canceling  a factor, — as  x in  the  first  divisor, — which  is  found  in  only 
one  of  them  ; nor  by  introducing  a new  factor , — as  in  multiplying  the 
first  dividend  by  4, — into  only  one  of  them, — since  these  expedients  do 
not  interfere  with  the  original  common  factors. 

The  suppression  of  a factor,  as  x in  this  example,  which  is  peculiar 
to  the  divisor,  is  necessary  ; for,  otherwise,  the  dividend  must  be  mul- 
tiplied by  this  factor  to  render  its  first  term  divisible, — and  this  would 
introduce  a new  common  factor  into  the  two  polynomials,  and  thus 
increase  the  greatest  common  measure,  (64). 


In  the  preceding  Example,  we  might  have  taken  the  first  remain 
der,  3 a2x-\~ax2 — 4a:3,  for  a divisor,  and  the  first  divisor  for  the  (livi 
dend.  The  operation  from  that  point  would  be  as  follows  ; — 


oa2  x-\-ax?  — 4.r3 
3 a2  -\-ax  — 4a:2 


4a2 — 5ax-\-  x2 
3 

12a2  — 15«a:-i-  3.r2  ( 4 
12a2  + 4aa" — 16a:2 

— 19aa:-(- 1 9a-2 


— 19aa:+  19a:2 
— • a + x 


3a2  + ax — 4a:2  ( — 3a — 4ar 
3a2  — 3 ax 


4 ax — 4a:2 
4aa: — 4a:2 


COMMON  MEASURE. 


3 

The  greatest  common  measure  is  here  found  to  be  — a-\-x,  which 
is  the  one.  before  determined,  with  its  signs  changed. 

The  effect  of  changing  the  signs  in  the  divisor,  would  only  be  to 
change  the  signs  in  the  quotient  resulting  from  the  division.  Hence 

(67.)  A Common  measure  of  two  or  more  quantities  may  have  all 
its  signs  changed,  without  ceasing  to  be  a common  measure  of  those 
quantities. 


The  Rule  which  has  been  given,  directs  that  one  of  the  two  quan- 
tities be  divided  by  the  other,  &c.,  without  distinguishing  the  divisor 
from  the  dividend.  We  therefore  remark  here,  that  the  same  common 
measure  will  be  found  by  taking  either  of  the  two  quantities  for  the 
divisor,  and  the  other  for  the  dividend. 


If  the  divisor  and  dividend  in  the  preceding  Example,  be  inter- 
changed with  each  other,  the  operation  will  be  as  follows  ; — 


3a3  — 3a2*-|-a* 2 — a:3 


4a2* — 5a*2  -f-  x3 

3 a 

12a3* — 15a2*2  + 3a*3  (4* 

12a3* — 12a2*2  +4a*3  — 4*4 


— 3a2*2  — a*3  -|-4*4 

— 3a2  — a*  +4*2 


3a2*2 — a*3-)-4*4 


1 3a3 — 3 a2*+  a*2 — *3  ( — a 

3a3+  a2* — 4a*2 

— 4a2*+  5a*2 — *3 

3 


— 12a2*+15a*2—  3*3  ( 4a: 

— 12a2* — 4a*2  + 16*3 


'9a*2 — 19*3 


a — * 


19a*2  — 19*3 


— 3a2 — a*+4*2  ( — 3a  + 4* 

— 3a2  + 3a* 


— 4a*+4*2 
— 4a*+4*2 
C 


36 


COMMON  MEASURE. 


In  this  case  the  factor  x is  contained  in  each  term  of  the  dividend, 
without  being  contained  in  each  term  of  the  divisor.  This  factor, 
therefore,  does  not  enter  into  the  composition  of  the  greatest  common 
measure,  (64),  and  might  he  canceled  before  dividing  ; and  this  would 
simplify  the  operation.  Hence 

(68.)  When  any  factor  is  common  to  all  the  terms  of  the  dividend , 
and  not  to  those  of  the  divisor,  such  factor  may  be  suppressed,  without 
affecting  the  greatest  common  measure  of  the  two  quantities 

It  follows  also  from  proposition  (64),  that 

(69.)  When  the  same  factor  is  contained  in  all  the  terms  of  two 
polynomials,  their  greatest  common  measure  may  be  found  by  multi- 
plying this  factor  into  the  greatest  common  measure  found  for  the  po- 
lynomials without  this  factor. 

For  example,  to  find  the  greatest  common  measure  of 
a2x  — x3  and  a2x-\-ax2 — 2x3. 

The  factor  x is  contained  in  all  the  terms,  and  the  two  polynomials 
without  this  factor  are  a 2 — x2  and  a2  -\-ax — 2x2. 

The  greatest  common  measure  of  these  will  be  found  to  be  a-i ; 
then,  mu1'. plying  x into  a — x,  we  find  ax — x2  for  the  greatest  common 
measure  of  the  given  polynomials. 


EXERCISES. 


1 . Find  the  greatest  common  measure  of 

a3—x3  and  ai—xi. 

2.  Find  the  greatest  common  measure  of 

a 2 — ax — 2x2  and  a2 — 3ax-f-2x2. 

3.  Find  the  greatest  common  measure  of 

3a2 — 2a — 1 and  4a3 — 2a2 — 3a+l. 

4.  Find  the  greatest  common  measure  of 

x2  + 2 ax-j-a2  and  x3 — a2x. 

5.  Find  the  greatest  common  measure  of 

2x3  — 16s  — 6 and  3s3— 24x— 9. 


Ans.  a— x 
Ans.  a—2x* 
Ans.  a— 1 i 


Ans.  x+a 


Hns.  x3  — Sx— 2 


COMMON  MULTIPLE. 


37 


6.  Find  the  greatest  common  measure  of 

a2 — 5ax-\-4x2  and  a3 — a2x-\-2>ax 2 — 3a:3. 

Ans.  a—x. 

7.  Find  the  greatest  common  measure  of 

a2 — 2 ax-\-x2  and  a3 — a2x — ax2-\-x3. 

Ans.  a2  -2ax-srx2 . 

8.  Find  the  greatest  common  measure  of 

6a2  + 7aa; — 3a;2  and  6a2  + llax-j-3x2 . 

Ans.  2a+3ar. 

9.  Find  the  greatest  common  measure  of 

lx2 — 23xy-\-Gy 2 and  5x3  — \8x2y-\-  Wxy2 — Gy3. 

Ans.  x — 3 y. 

10.  Find  the  greatest  common  measure  of 

a4-\-a2y2-\-y4  and  a3 — 2a2y-\-2ay2 — ys. 

Ans.  a2 — ay-\-y2. 

11.  Find  the  greatest  common  measure  of 

3a5  + 6a4a;+3a3a:2  and  a3z~l-3a2x2  ~{-3ax3  -\~x4. 

Ans.  a2  + 2ax-{-x2 . 

12.  Find  the  greatest  common  measure  of 

a:3—  ax2 — 8a2x-\-Ga3  and  x 4 — 3ax3 — 8a2x2 -\-18a3x  — 8a4 . 

Ans.  x2  + 2ax — 2a3. 


COMMON  MULTIPLE. 

(70.)  One  quantity  is  called  a multiple  of  another,  if  it  can  be 
neasured  by  the  other,  that  is,  divided  by  it,  without  a remainder ; 
fmd 

A common  multiple  of  two  or  more  quantities  is  any  quantity  that 
■an  be  divided  by  each  of  them,  without  a remainder. 

Thus  8a3a:2  is  a common  multiple  of  2a2a:  and  4a;2. 

CCP*  Name  a Common  Multiple  of  3 ab  and  5c 2 . Of  ax2  and  2 ay. 
)f  3a2  and  4a;2.  Of  5bc  and  a;2.  Of  y3  and  5a:3. 

Least  Common  Multiple. 

(71.)  The  least  common  multiple  of  two  or  more  quantities,  is  th< 
mailest  quantity  that  can  be  divided  by  each  of  them,  without  a re 
rainder. 

Thus  6aa:2  is  the  least  common  multiple  of  6a  and  6a;2. 


38 


COMMON  MULTIPLE. 


tO55’  What  is  the  Least  Common  Multiple  of  2 ab  and  3b2  ? Of 
4x  and  4 y2  1 Of  5a  and  3 a2b1  Of  ab  and  2 be2  1 Of  7 ax  and  ax 2 ] 
Of  3a2  and  6 ay2  1 Of  axy  and  5x2y 2 ? 

(72.)  The  least  common  multiple  of  two  or  more  quantities  is  com- 
posed of  the  smallest  selection  of  factors  that  includes  the  factors  of 
each  given  quantity. 

For  example,  take  the  quantities  3 ab2c  and  6a2bxy, 

Resolving  these  two  quantities  into  their  prime  factors , we  have 
3 abbe  and  3x2  aab  xy. 

If  we  take  3x2  aa  bbc  xy,  we  shall  have  the  smallest  selection  of 
factors  that  includes  the  factors  of  each  of  the  two  given  quantities. 

Then  the  product  &a2b2cxy  is  the  least  common  multiple,  because 
it  is  the  smallest  quantity  that  each  of  the  given  quantities  will  divide, 
without  a remainder. 

(73.)  The  least  common  multiple  of  two  quantities,  is  equal  to 
their  product  divided  by  their  greatest  common  measure. 

For  since  the  greatest  common  measure  of  two  quantities,  is  com- 
posed of  all  the  factors  which  are  common  to  those  quantities,  (64), 
these  factors  will  enter  twice  in  the  product  of  the  quantities. 

If,  therefore,  the  product  be  divided  by  the  greatest  common  mea- 
sure, the  quotient  will  contain  only  those  factors  which  are  conunon  to 
the  two  quantities,  and  those  which  dice  peculiar  to  each  of  them  ; and 
these  are  the  factors  of  the  least  common  multiple,  (72). 


RULE  VIII. 

(74.)  To  find  the  Least  Common  Multiple  of  Two  or  more 
Quantities. 

1.  Set  the  quantities  in  a line,  from  left  to  right,  and  divide  any 
hvo  or  more  of  them  by  any  prime  quantity,  greater  than  unity,  thal 
will  divide  them,  without  a remainder, — placing  the  quotients  and  the  1 
undivided  quantities  in  a line  below. 

2.  Divide  any  two  or  more  of  the  quantities  in  the  lower  line,  a-  . 
before  ; and  so  on,  until  no  two  quantities  in  the  lowest  line  can  be  ; 
divided.  The  product  of  the  divisors  and  quantities  in  the 

line,  will  he  the  least  common  multiple  required. 

3.  If  no  two  of  the  given  quantities  can  be  divided  as  above,  th- 
product,  of  all  the  quantities  will  be  their  least  common  multiple. 


COMMON  MULTIPLE. 


39 


EXAMPLE. 

R 

To  find  the  least  common  multiple  of 

2ax2,  5a2y,  and  3 y — 4 y3 


3 J 3ax2 

6 a2y 

3y — 4 y3 

aj  ax2 

2 a2y 

3y—iy3 

y) 

2 ay 

3 y—fy3 

X 2 

2 a 

3-4  y2 

In  the  first  operation  we  divide  3 ax2  and  6 a2y  hy  3,  and  set  down 
3 y — 4 y3  without  dividing  it ; and  in  like  manner  in  the  second  opera- 
tion. In  the  third  operation  we  set  down  x2  without  di\riding  it. 

Then  3 ayx2  x2ax(3 — 4 y2),  equal  to  18 a2x2y — 24 a2x2y3,  is  the 
.east  common  multiple  of  the  three  given  quantities. 

1 

JThis  Rule  depends  on  proposition  (72) : the  divisors  and  quantities 
n the  lowest  line,  are  the  smallest  selection  of  factors  that  includes  the 
factors  of  each  given  quantity. 

EXERCISES. 


1.  Find  the  least  common  multiple  of 

ax2,  2a2 y,  4 y-{-y2,  and  ax2 4a2, 

Ans.  8 a3x2y  + 2a3x2y2  -\-3>2a2x2y-\-8a2x2y2 . 

2.  Find  the  least  common  multiple  of 

2 ay2,  Aay2,  2x  — 4a:2,  and  a:3  + <z;r2- 

Ans.  £ax3y2  — 8axiy 2 -f-4 a2x2y2  — 8a2x3y2, 


I 

4' 


1 


Find  the  least  common  multiple  of 

3a2,  ax2,  3a+6a2,  and  x3 — 3a:2. 

Ans.  3a2x3  -\-&a3x3 — 9a2a:2 — 18a3a;2. 

Find  the  least  common  multiple  of 

4 y3,  2 ay2,  5a  — 5ab,  and  10a — 5. 

Ans.  ^5a2tj3  — 40 a2 by3  — 20 a.y3  -f  20 aby3. 

Find  the  least  common  multiple  of 

5a,  10 ab,  3y+3y2,  and  6y3+3y2. 

Ans.  90aby3  -i-QOaby*  + 30aby2. 


40  COMMON  MULTIPLE. 

6.  Find  the  least  common  multiple  of 

4,  4x2,  8y — 8,  and  2x3 — ax 3. 

Ans.  16  x3y — 16x3  — 8ax3  y-\-8ax3 

7.  Find  the  least  common  multiple  of 

10,  5ax,  4 a — 8 y2,  and  2x+6x2. 

Ans.  20 a2x — 40axr/2  + 60a2x2  — 120ax2y2 

8.  Find  the  least  common  multiple  of 

3 y2 , 12 y,  5x 2 — 10,  and  4 y — 8 y3. 

Ans.  60x2?/2  — 120i/2 — 120x2t/4-)-240?/4 

9 Find  the  least  common  multiple  of 

14,  la,  2 y2 — 4,  and  28-\-lay. 

Ans.  56ay2-t-112a-t-14a2y3 — 28 a2y. 


10.  Find  the  least  common  multiple  of 

a4 — x4,  and  a3 — a2x — ax2-)-*3,  (73). 

Ans.  a5 — a4x — ax4+x5. 

11.  Find  the  least  common  multiple  of 

x2 -\-2bx-\-b2  and  x3—b2x.  (73). 

Ans.  x4  + 5a;3 — b2x2 — b3x. 

12.  Find  the  least  common  multiple  of 

a2  -3ab-\-2b2  and  a2—  ab — 2b2.  (73). 

Ans.  a3 — 2a2b-ab2-\-2b3. 

13.. Find  the  least  common  multiple  of 

4a3  — 2a2—  3a+l  and  3a2—  2a— 1.  (73). 

Ans.  12  a4 — 2a3  — lla2  + l. 


41 


CHAPTER  IV. 

FRACTIONS. 


(75.)  An  algebraic  fraction  represents  the  Quotient  of  its  nume- 
rator divided  by  its  denominator. 

Thus  represents  the  Quotient  of  a divided  by  3x. 

In  reading  an  algebraic  Fraction,  it  will  often  be  necessary  to  use 

tthe  terms  numerator  and  denominator , to  avoid  ambiguity  in  refer- 
ence to  the  division  which  is  expressed. 

Thus  if  the  Fraction 


a-\-x 

y 

be  read,  a-\-x  divided  by  y,  it  might  be  understood  that  only  x is  to  be 
divided  by  y.  But  the  true  sense  would  be  conveyed  by  saying,  nume- 
rator a+x,  denominator  y. 

A Fraction  is  thus  employed  to  represent  the  Quotient,  when  the 
divisor  is  not  a factor  of  the  dividend.  The  quotient  in  this  case  may 
also  be  represented  by  means  of 

Negative  Exponents. 


(7G.)  Any  quantity  with  a negative  exponent  is  equivalent  to  a 
unit  divided  by  the  same  quantity  with  the  sign  of  its  exponent 
changed. 


Thus  ar2,  a with  exponent  — 2,  is  equivalent  to 


For  a2a  2 is  equal  to  a0,  (41) ; 
and  by  dividing  each  of  these  equals  by  a2, 

Q,®  1 

we  find  ar2  equal  to  — r or  — , (48). 

a2  a- 


0^“  What  is  the  fractional  equivalent  of  a~3  l and  how  is  it 
proved  1 Of  x_i  ? and  how  is  it  proved  1 Of  y~5  1 and  how  is  it 
proved!  Of  ( a-\-x)~ 2 • Of  ( x — y)~3  ? 

3* 


42 


FRACTIONS. 


(77.)  When  the  Divisor  is  not  a factor  of  the  dividend,  the  Quotient 
may  he  represented  by  a fraction  (75),  or  by  the  dividend  multiplied, 
into  the  divisor  with  the  sign  of  its  exponent  changed. 


a — x2  will  give  the  quotient  ax~2 , because  this  quotient  multiplied 
into  the  divisor  x2,  produces  ax°,  (41),  which  is  equal  to  a,  the  divi- 
dend, (48). 


We  have,  therefore,  ax  2 for  an  i?itegral,  and  — for  a fractional 


form  of  the  quotient  of  a^x2 . 


US?"  "What  is  the  integral  form  of  the  Quotient  of  a — S3  ? and 
how  is  it  proved?  Of  x^-y*  1 and  how  is  it  proved?  Of  1-^a2  ? 
and  how  is  it  proved  ? Ofa2-i-a:?  Of  b — ac2!  Of  a — 5? 

Transfer  of  Factors. 


(78.)  Any  factor  may  he  transferred  from  the  denominator  to  the 
numerator,  and  vice  versa , by  changing  the  sign  of  its  exponent. 

For  example,  if  we  divide  a by  x2y,  fractionally, 


we  have  a^x2y  equal  to  — — . 

x2y 


If  we  divide  a by  the  factors  x2  and  y,  separately,  we  shall  find 

CiX  ^ 

a —r x2  equal  to  ax~2 , (77),  and  ax~2  ~y  equal  to . 

Hence  a divided  by  x2y  gives 


a ax  2 

— or , 

x2y  y 

These  two  quotients  being  necessarily  equal  to  each  other,  we  see 
that  x2  may  be  transferred  from  the  denominator  to  the  numerator,  by 
changing  the  sign  of  its  exponent. 

If  we  also  transfer  the  factor  y,  we  shall  have 

a ax~ 2y~x  _ , , . 

— py-  equal  to  — - or  ax  -y  1,  (77). 


If  we  transfer  the  factor  a from  the  numerator  a , or  la,  to  the  de- 
nominator, we  find 


— - — equal  to 
x-y 


1 

a~1x2y 


!GP“  Under  what  different  forms  may  the  Quotient  of  a — bx2  be 
represented?  Of  3a^x2  ? Of  2c— 3a4?  Of  ab  — x2y2l 


FRACTIONS. 


43 


Reciprocals  of  Quantities. 


(79.)  The  reciprocal  of  a quantity  is  a unit  divided  by  the  quantity, 

Thus  the  reciprocal  of  a 3 is  — or  a~3  (76). 

(80.)  The  reciprooal  of  a Fraction  is  equivalent  to  the  fraction  in- 
verted; that  is,  with  its  numerator  and  denominator  taken  the  one  for 
ihe  other. 

CL  1 CC^ 

Thus  the  reciprocal  of  — , or  ax~2  (77),  is — , which  becomes  — - 

x2  ax  2 a 

by  transferring  x~2  from  the  denominator  to  the  numerator,  (78). 


DGP“  Under  what  different  forms  may  the  Reciprocal  of  — ■ he  ex- 

x 

pressed  ? The  reciprocal  of  — 2~  ? The  reciprocal  of  Of  —~p- ? 

?/  cy 


Constant  Value  of  a Fraction. 


(81.)  The  value  of  a Fraction  remains  the  same  when  its  numera- 
tor and  denominator  are  both  multiplied,  or  both  divided,  by  the  same 
quantity. 

For  example,  if  we  multiply  both  terms  of  ~ by  y, 

_ ,11  a . ay 
we  shall  have  — equal  to  — - — . 

x 2 x2y 

' For  the  first  of  these  fractions  is  equal  to  ax~ 2 ; 

and  the  second  is  equal  to  ax~2yy_1,  (77). 

But  yy~l  is  equal  to  y°,  (41),  and  by  cancelling  this  factor,  (48), 
we  find  the  second  fraction  also  equal  to  ax~2 . Hence  the  two  frac- 
tions are  equal  to  each  other. 

. a2  . a2x  c2  . , c2m2 

LGr3”  rrove  that  — is  equal  to  — — . I hat  — is  equal  to  — - — — , 

b bx  Xs  x3m2 

Signs  of  Fractions. 

(82.)  Since  a Fraction  represents  the  quotient  of  its  numerator  di- 
vided by  its  denominator,  a Fraction  is  positive  when  its  numerator 
and  denominator  have  the  same  sign,  and  negative  when  they  have 
contrary  signs,  (49). 

C* 


44 


FRACTIONS. 


OGP3  Say  whether  the  Fraction  — is  positive  or  negative.  Say 


whether 


is  positive  or  negative.  Say  whether 
— b — b 


is  positive  or 


i i —ax  ... 

negative.  Say  whether  — — — is  positive  or  negative. 


(83.)  The  sign  -ff  or  — prefixed  to  a Fraction , — not  the  sign  of 
either  the  numerator  or  denominator, — shows  whether  the  fraction 
enters  additively  or  siibtr  actively  into  a calculation. 


Thus  H — denotes  that  the  fraction  — -- — , which  is  in  itself 

b b 

negative,  (82),  is  to  be  added,  in  the  calculation  into  which  it  enters. 


When  no  sign  is  prefixed  to  a fraction,  -f-  is  always  understood. 


(84.)  The  signs  of  both  the  numerator  and  denominator  may  be 
changed,  or  the  sign  of  either  of  them  with  the  sign  prefixed  to  the 
Fraction, — without  affecting  the  value  of  the  fraction. 


Thus  — is  equivalent  to 
positive , (82). 


since  both  these  fractions  are 


. . Or  . . , Or 

Also  — is  equivalent  — — ■ — . 
b b 


For  the  fraction  — - — is  negative, 
b 


its  numerator  and  denominator  having  contrary  signs ; this  nega- 
tive fraction  becomes  positive,  when  subtracted,  as  required  by  the 
sign  — prefixed  to  it,  and  it  is  then  equivalent  to  the  first  fraction. 


A Polynomial  is  changed  from  positive  to  negative,  or  from  nega- 
tive to  positive,  by  changing  the  sign  of  each  of  its  tentis,  (38). 

For  example,  if  a — b is  positive,  a must  be  greater  than  b ; then 
changing  the  signs,  b — a will  be  negative. 

The  Fraction  — — - is  therefore  equivalent  to  - — — . 

x—y  y—x 

tCP"1  What  other  changes  may  be  made  in  the  signs,  without  at 
ecting  the  value,  of  this  fraction  ? 


FRACTIONS. 


45 


Fractions  Reduced  to  their  Lowest  Terms. 

(85.)  A Fraction  is  reduced  to  lower  terms  by  dividing  its  numera- 
tor and  denominator  by  the  same  common  measure.  This  simplifies 
the  fraction,  without  altering  its  value,  (81). 


A monomial  common  measure  may  usually  be  known  by  inspection. 
Thus  to  reduce  the  Fraction 

4.a2b 

6 ac — 8 a2x 

It  is  obvious  that  we  have  only  to  divide  its  numerator  and  deno- 
minator by  2a.  This  gives  us  the  equivalent  Fraction, 

2 ab 

3c — 4 ax ' 

A binomial  common  measure  may  often  be  discovered  from  the 
principles  which  have  been  established  for  the  decomposition  of  Poly- 
nomials. 

Thus  to  reduce  the  Fraction 

a2  —b2 
a2  -\-2ab-\-b2 

By  proposition  (57)  we  can  divide  the  numerator  by  a-\-b  ; and  by 
(59),  we  can  divide  the  denominator  by  the  same  quantity 
Thus  dividing  we  find  the  equivalent  Fraction 

a—b 

a-\-b 

In  all  cases  in  which  a Fraction  admits  of  being  reduced,  we  may 
apply 


RULE  IX. 


(86.)  To  Reduce  a Fraction  to  its  Lowest  Terms. 

Divide  the  numerator  and  denominator  by  their  greatest  common 
measure,  (66)  : the  quotients  will  be  the  lowest  terms  of  the  given 
fraction. 


1. 

2. 


Reduce 

Reduce 


3a*2 

3 a2b — 9a3 
2a2y  -\-\ay 
8 a 


EXERCISES. 

to  its  lowest  terms. 

2 

- to  its  lowest  terms. 


Ans. 

Am. 


x2 

ab  — 3a2 
ay+2y2 

4 


46 


r FACTIONS. 


„ , a2-b2  . , 

3 Reduce  — — — to  its  lowest  terms.  (57). 
a6-\-bA 


a4  —x4 

4.  Reduce  — — — to  its  lowest  terms. 

ab  —aAx2 


^2 

£>.  Reduce  — to  its  lowest  terms. 


Ans. 


a — b 


a2  — ab-\-b2 
a2  + a:2 


Ans 


x2  +2  ax+a2 


Ans. 


6.  Reduce 


7.  Reduce 


a3 — ay2 
a2  -\-2ay-\-y2 


x—a 
x+a ' 


to  its  lowest  terms. 


Ans. 


a2  —ay 
a+y 


a * —x“ 


a 3 — 2>a2x-\-2>az2  — x3 


to  its  lowest  terms. 


Ans. 


a-\-x 


2 ax2—a2x — a3 

8.  Reduce — to  its  lowest  terms. 

2x2  + 3ax+a 2 


a3  + 2a2x-j-3a2x2 

y.  Reduce  — — — to  its  lowest  terms. 


2a4  —3a3  x — 5a2x2 


Ans. 


a2  — 2ax-\-x2  ’ 


. ax — a2 

Ans. 

x-\-a 


a,-\-2x-\-3x2 


2 a2  —oax  — 5x2  ' 


, _ ,,  . 6a2+7aa? — 3x2  . 

10.  Reduce  — — - to  its  lowest  terms. 

6a2  + llaz+3z2 


Ans. 


6a— z 


a ^ — |-  a ^ x ^ 'X/  ^ 

11.  Reduce  — — — — 7 to  its  lowest  terms. 


a4  -{-a3x— ax3  — x4 


3a  + ce 
a2  —ax+x2 


a“ — x“ 

t*  i 5a5  + 10a4w+5«3w2  . 

1 2.  Reduce  — to  its  lowest  terms. 

a3y-\-  2 a2y2  + 2 ay3  -\-y4 

5a4-\-5a3y 


A?is. 


a2y+ay2+y3 


FRACTIONS. 


47 


Fractions  Reduced  to  a Common  Denominator. 

(87.)  Two  or  more  Fractions  are  said  to  have  a common  denomina- 
tor, when  they  have  the  same  quantity  for  a denominator. 


Thus 


and have  a common  denominator. 

a-\-b  a-\-b 


Two  or  more  Fractions  may  often  be  reduced,  very  readily,  to  a 
common  denominator,  by  multiplying  both  the  numerator  and  denomi- 
nator of  one  or  more  of  them,  so  as  to  make  the  denominator  the  same 
for  each. 

For  example,  to  reduce  to  a common  denominator  the  Fractions 

a , be 

and  , 

2 a — 2x  '3a  — 3x 


We  have  only  to  multiply  the  terms  of  the  first  fraction  by  3,  and 
those  of  the  second  by  2.  This  gives  the  equivalent  Fractions 


3 a 

6 a — 6x 


and 


2 be 

6 a — 6x 


(81). 


Observe  that  this  reduction  does  not  alter  the  values  of  the  given 
Fractions. 

When  this  method  cannot  be  obviously  applied,  we  adopt 


RULE  X. 

(88.)  To  Reduce  tivo  or  more  Fractions  to  a Common  Denomi- 
nator. 


1.  Multiply  each  numerator  by  all  the  denominators  except  its 
oicn,  for  new  numerators  ; and  multiply  all  the  denominators  together, 
for  a common  denominator . 

2.  If  the  Least  Common  Denominator  he  required, — Find  the 
least  common  multiple  of  the  given  denominators,  for  the  Common 
denominator.  Divide  this  Multiple  by  the  denominator  of  each  given 
Fraction,  and  multiply  the  quotient  by  the  numerator,  for  the  new 
numerators. 


EXAMPLES. 


1.  To  reduce  — , — , and  - — - to  a common  denominator. 
3x  G y-\-  2 

For  the  new  numerators,  we  have 

a . 6 . (y-\-  2),  equal  to  &ay-\- 1 2 a ; 
b.3x.(y+2),  “ 3bxy-\-6bx; 

and  c . 3x  . 6 “ 18ca:. 

A.nd  the  common  denominator  is 

3x  . 6 . (y-\-  2),  equal  to  18a;?/+36;2. 


48 


FRACTIONS. 


The  given  Fractions  are  thus  reduced  to 
6aw+12a  3bxy-\-§bx  18  cx 
I8xy+3§x  ’ lQxy+3&x  ’ \Qxy+3bx  ’ resPectlve 

2.  To  reduce  the  same  Fractions  to  the  least  common  denominator. 

The  least  common  multiple  of  the  denominators  3x,  6,  and  y-\-  2 
will  he  found  to  be  bxy-\~\2x , (74),  which  is  the  required  denominator. 

Dividing  this  Multiple  by  each  given  denominator,  and  multiplying 
the  quotients  by  the  given  numerators,  respectively,  we  find  the  new 
numerators, 

2ay-\-4a,  bxy-\-2bx,  and  6 cx. 

The  Fractions  reduced  to  their  least  common  denominator,  are  then 
2ay-\-4a  bxy-\-2bx  6 cx 

d>xy+ I2x  ’ Qxy+\2x  ’ 6xy+l2x  ’ resPectlvely- 

Each  of  the  given  Fractions  should  be  in  its  lowest  terms,  before 
proceeding  to  find  their  least  common  denominator ; otherwise,  the  de- 
nominator found  will  not,  in  all  cases,  be  the  smallest  by  means  of 
which  the  values  of  the  several  Fractions  may  be  expressed. 

In  finding  a Common  Denominator  as  above,  the  numerator  and 
denominator  of  each  given  fraction  are  multiplied  by  the  same 
quantity. 

Thus  in  the  first  example,  — has  both  its  terms  multiplied  by  6 

oCC 

and  y-\- 2, — producing  the  new  terms  bay-\-\2a  and  18xy+36x. 

Hence  the  values  of  the  given  fractions  are  not  altered  in  reducing 
them  to  a common  denominator,  (81). 


EXERCISES. 


1.  Reduce  — , 

O 


b 

2x 


and to  a common  denominator. 

x— 4 


2.  Reduce  — 
ai 


n 2 ax2 — 8 ax  3bx— 125  6cz 
,lS'  6x2—  24x  ' 6x2  — 24x  ’ 6x2—24x 
a a-\-c 

, and to  a common  denominator. 

y 2 i —y 

2y2  — 2y2  a2—a3y  azy2  -\-a2cy2 
a2y2—a2y%  a2y2  —a2y%  ’ a2y2—a2i/a 


FRACTIONS. 


49 


3.  Reduce 


* ’ 


C , (2+5 

, and  — — to  a common  denommator. 


Ans. 


ax  a b — 1 

i.  Reduce  — , — , and to  a common  denommator. 

2 3a:  l + » 


_ a3  b a-\-b 

5.  Reduce  — , — - , and 

y 2x2  xy 


Ans. 

„ _ -i  a b ab 

6.  Reduce  — - , - , and  -5— — - 
2x 2 y x2-{-x3 


Ans. 


3a2y3  cxy 2 3axy-\-\5xy 

3xy3  ' 3 xy3  ’ 3 xy3 

mon  denommator. 

3ax2  A^abx2  2a-\-2ab  6 bx — 6x 

6x-j-6bx  ’ 6z+6 bx  ’ 6a:+6 bx 

to  a common  denominator. 

2 a3x3y  bxy2  2ax2yJr2bx2y 
2 x3y2  ’ 2x3y 2 ' 2x 3y2 

to  the  least  common  denominator. 

ay-\-axy  2bx2-\-2bx3  2 aby 

2x2y-\-2x3y  ’ 2x2y-\-2x3y  ’ 2x2y-\-2x3y 


7 Reduce  — , — , and  — to  the  least  common  denominator. 

4 3 y 6 y2 

3 x2y2  Aay  2 a — 2x 

Am‘  Ity2  ’ l2p  ’ 


0 _ , a b c 

8.  Reduce  — , - , and  — 

yi  5 yh—y2 

Ans. 


to  the  least  common  denominator. 
5ay — 5a  by3 — by2  5c 


5y3 — 5y2  ’ 5 y3 — 5y2  ’ 5y3 — 5y2 


'll  CL  (lA 

9.  Reduce  - . — ^ , and to  the  least  common  denominator. 

4 2x2  2 + x 


Ans. 


2 x2y-\-x3y  4a  + 2aa;  4 a2x2 


rjQ 

10.  Reduce  — , — , and 


3 ’ 3 y 


Ans. 


8a:2  + 4z3  ’ 822+4£3’  8x2+4x3  ' 


to  the  least  common  denomi- 


6y—3y2 


2 xy — xy2 — 2 y-\-y2  2a2 — a2y 


&y—3y2 


5y  — 3y2  ’ 6y-3y2 


nator. 


50 


FRACTIONS. 


Integral  and  Mixed  Quantities  Reduced  to  Improper 
Fractions. 

(89.)  An  integral  quantity  is  one  which  does  not  contain  any 
fractional  expression;  as  3 ax2,  or  2ab—5xy. 

(90.)  A mixed  quantity  is  partly  integral  and  partly  fractional ; 

CL  (C 

as  Sax2 -\ , or  2 ab . 

0 c 1+2/ 

(91.)  An  improper  Fraction  is  a fraction  whose  value  may  he  ex- 
pressed by  an  integral  or  a mixed  quantity. 

5 

Thus  — is  an  improper  fraction  whose  value  is  2 a-\ . 


RULE  XI. 

(92.)  To  Reduce  an  Integral  or  a Mixed  Quantity  to  an  Impro- 
per Fraction. 

1.  Under  an  integral  quantity7,  regarded  as  a numerator,  set  1 for 
a denominator.  Or  multiply  the  integral  quantity7  by  any  proposed 
denominator  ; the  product  will  be  the  numerator. 

2.  In  a mixed  quantity,  multiply  the  integral  part  by  the  denomi- 
nator annexed  ; add  the  numeratoi-  to  the  product  when  the  sign  be- 
fore the  fractional  part  is  +,  but  subtract  the  numerator  when  this 
sign  is  — ; and  place  the  result  over  said  denominator. 

EXAMPLES. 


1.  To  reduce  3a2  to  a Fraction  whose  denominator  shall  be  a — 2x. 
3 a2 

3 a2  is  the  same  as  — — ; and  by  multiplying  both  terms  of  this 
fraction  by  the  proposed  denominator,  we  have 


3a2  equal  to 


3a3  — 6a2 x 
a.  — 2x 


, (81). 


2. 


To  reduce  3a2 


to  an  improper  Fraction 


Multiplying  3a2  by  the  denominator  2,  and  subtracting  the  nume- 
rator a — x2  from  the  product, — observing  to  change  the  sigfis  of  the 
'uimerator,  (36), — we  have 


3a2- 


a — x‘ 


equal  to 


6a2  — a-\-x2 


2 


2 


FRACTIONS. 


51 


The  reason  of  this  operation  will  be  evident,  if  we  consider  that, 
by  multiplying  the  oa2  by  2,  we  reduce  the  integral  part  to  a common 
denominator  with  the  fraction  annexed  to  it,  according  to  the  first  part 
of  the  Rule.  The  operation  then  consists  in  subtracting  numerator 
from  numerator,  and  placing  their  difference  over  the  common  denomi- 
nator. 

Another  view  to  be  taken  of  the  preceding  operations,  is,  that  the 
integral  term  3a2  is  multiplied,  and  the  product  is  then  taken  to  be 
divided,  by  the  same  quantity,  namely,  the  denominator.  The  value 
therefore  remains  the  same. 

EXERCISES. 


1.  Reduce  4 ax2  to  a Fraction  whose  denominator  shall  be  y-{-  2. 

Aax2y-\-8ax2 


Ans. 


y+ 2 


2 Reduce  a2 -\- 3x  to  a Fraction  whose  denominator  shall  be  2 y2. 

2a2y2  -{-6xy2 


Am. 


2y2 


3.  Reduce  5xy  to  a Fraction  whose  denominator  shall  be  3 + x2 . 

\Sxy-\-  5x*y 


Ans. 


3 + z2 


4.  Reduce  y2 —5  to  a Fraction  whose  denominator  shall  be  3ax2. 

. 3 ax2y2 — \5ax2 

Am-  — m — 

5.  Reduce  a+5  to  a Fraction  whose  denominator  shall  be  a — b. 

a2-b 1 


Ans. 


a — b 


6.  Reduce  Say2  to  a Fraction  whose  denominator  shall  be  a 2 — b2 

^ 5a3y2—5ab2y2 

a2 —b2 

7 Reduce  *2-J-l  to  a Fraction  whose  denominator  shall  be  3 a2y. 

Ans.  3a2*2?/+3*2? 
3 a2y 

8 Reduce  ab2 — x to  a Fraction  whose  denominator  shall  be  1 — y 

. ab2 — x—ab2y-\-xi 
Ans.  — — - 


52 


FRACTIONS. 


9.  Reduce  a2  + #+-  to  an  improper  Fraction. 


Am.  a2l±^±l. 

y 


10.  Reduce  ax2 to  an  improper  Fraction. 

y+ 1 


a+x 


Am. 

y+ 1 


11.  Reduce  5a +35 — to  an  improper  Fraction. 

o 

Ans. 


1 4 a -f~  9 b — x 


12.  Reduce  2a— 4c-| to  an  improper  Fraction. 

Ans. 

1 y2 

13.  Reduce  2-\-y2  — — — — to  an  improper  Fraction. 


11a — 20c — y 2 


Atis. 


7+5y 2 


/£  3 

14.  Reduce  a2-)-#2 — — to  an  improper  Fraction. 


a—x 


' Ans. 


ax 2 —a2x 


a — x 


q2 CLCC 

15.  Reduce  a — x-\ to  an  improper  Fraction. 


Ans. 


a2—x 2 


4iT — 4 

16.  Reduce  1 + 2z — — to  an  improper  Fraction. 


5a; 


5y2 


oi«S. 


z+lOx2-1  4 
5x 


17.  Reduce  2x-f  2 y-\-~ — to  an  improper  Fraction. 
/wX — oy 


Ans. 


4x2— 2 xy—y2 


18.  Reduce  a-\-x — — — - — ; — , to  an  improper  Fraction. 


2x — oy 


a2  -ax-\-x2 


Arts. 


2a-3 


a2— ax- fa;2 


FRACTIONS. 


53 


Improper  Fractions  Reduced  to  Integral  or  Mixed 
Quantities. 

By  reversing  the  preceding  Rule,  we  have 
RULE  XII. 

(93.)  To  Reduce  an  Improper  Fraction  to  an  Integral  or  a 

Mixed  Quantity. 

Divide  the  numerator  by  the  denominator  for  the  integral  part, 
and  set  the  denominator  under  the  remainder,  if  any,  for  the  fractional 
part,  of  the  result.  Connect  the  fractional  to  the  integral  part  by  the 
sign  + ; or  change  the  sign  of  the  numerator  or  denominator,  and 
connect  it  by  the  sign  — . 

EXAMPLE. 

To  reduce  to  an  integral  or  a mixed  quantity  the  Fraction 

9a 2 c + 3ab  — 2b + y 
3 a 

Dividing  the  numerator  by  the  denominator,  we  find  the  integral  quo- 
tient to  be  3 ac-\-b,  and  the  remainder  — 2 b-\-y. 

Setting  the  denominater  under  this  remainder,  and  connecting  the 
fraction  so  formed  to  the  integral  quotient,  by  the  sign  +,  the  result  is 

3 ac+b+-^±Z  . 

Or,  changing  the  signs  in  the  numerator  — 2 b-\-y,  and  connecting 
the  fraction  by  the  sign  — , the  result,  under  a somewhat  simpler 
form, 

• o 7 2b— y 

is  3 ac-\-b — — . 

3 a 

This  form  is  simpler  than  the  preceding,  as  it  dispenses  with  one 
sign  in  the  numerator  of  the  fractional  part. 

The  reason  of  the  preceding  operation  is  evident  from  the  conside- 
ration that  every  Fraction  is  equal  to  its  numerator  divided  by  its  de- 
nominator, (75).  After  obtaining  the  quotient  3 ac-\-b, — the  divisor 
3a  not  being  contained  in  the  remainder  — 2 bfy,  the  division  of  these 
terms  is  indicated  by  setting  the  divisor  under  them. 

The  fractional  part  of  the  result  must  evidently  he  added  to  the 
integral  part ; and  this  addition  is  indicated  by  placing  the  sign  + 
before  the  fraction.  But  the  value  of  the  fraction  annexed,  will  not 
be  affected  by  changing  the  sign  -f-  before  it  to  — , if  at  the  same  time 
we  change  the  signs  in  the  numerator,  (84). 


54 


FRACTIONS. 


The  Fraction  formed  of  the  divisor  and  remainder,  •wall  he  in  its 
lowest  terms , or  not,  according  as  the  improper  fraction  reduced,  is,  or 
is  not,  in  its  lowest  terms. 

For,  if  the  dividend  and  divisor  have  any  common  measure , the 
divisor  and  remainder  will  have  the  same  common  measure,  (65). 


EXERCISES. 

^3 <^3 

1 . Reduce to  an  integral  or  a mixed  quantity. 

a—x 

Ans.  a2  +ax-\-x2 . 

IQx2 5x-\-3 

2.  Reduce — to  an  integral  or  a mixed  quantity. 

OX 


Ans.  2x— 1+—  . 

ox 


4 

3.  Reduce to  an  integral  or  a mixed  quantity. 

a—y 

A?is.  a3  -\-a2y-\-ay2  +t/3 

£p4  — 3x^7/^  | " CLX 

4.  Reduce  ^ = — — to  an  integral  or  a mixed  quantity. 

x*  — Siy1 

_ ax 

Ans.  z2  + -v — — , • 

^ 4a2x2  — 3ay—  2b  . . 

5.  Reduce  ' to  an  integral  or  a mixed  quantity. 


2a2 


Ans.  2x2- 


x2  — 3 y2 
tntity. 
3(2?/+  2b 


9ax2  — 2x+3  . . 

b.  Reduce to  an  integral  or  a mixed  quantity. 


2a2 


6a 


Ans.  3x2 - 


2x—3 


I yi3 

7.  Reduce — - — to  an  integral  or  a mixed  quantity. 


a+y 


Ans.  a2  —ay Ay"1  ■ 


$2  y2  — 4: 

8.  Reduce to  an  integral  or  a mixed  quantity. 

x+y 

Ans.  x — y- 

Q-d'ij — - j- 

9.  Reduce  — — - — to  an  integral  or  a mixed  quantity’. 

2 a-y  B H 

„ Ans.  2 y+- 


x+y 


:a  —y 


FRACTIONS. 


55 


ADDITION  OF  FRACTIONS- 


(94.)  The  Sum  of  two  or  more  Fractions  is  found  by  means  of  a 
common  denominator. 

mi  i o ra  t b . a-\-b 
1 hus  the  Sum  oi  - and  - is  — — - . 

xxx 

For  it  is  evident  that  a divided  by  x,  added  to  b divided  by  x, 
makes  the  sum  of  a and  b divided  by  x. 

In  other  words,  if  each  of  the  ‘parts  a and  b he  divided  by  x, 
the  whole  a-\-b  will  be  divided  by  x. 

Hence  we  have 

RULE  XIII. 

(95.)  For  the  Addition  of  Fractions. 


1 . If  the  fractions  have  not  a common  denominator , reduce  them 
to  a common  denominator. 

2.  Add  the  numerators  together,  and  place  the  Sum,  as  a numera- 
tor, over  their  common  denominator. 

3.  Mixed  quantities  may  be  added  under  the  form  of  improper 
fractions;  or  the  integral  and  the  fractional  parts  may  be  added  sepa- 
rately. 


EXAMPLE. 


b c 

To  add  together  2 a-\ — and  3 a . 

x y 

Reducing  these  mixed  quantities  to  improper  fractions,  they  be 
come 

2 ax-\-b  , oay — c 

— and  — - — - . 

x y 

Reducing  these  fractions  to  a common  denominator,  we  have 
2axy-\-bij  3 axy — cx 

xy  xy 

Placing  the  sum  of  these  numerators  over  the  common  denomma 
tor,  the  result  is, 

5axyfby—cx  by — cx 

, equal  to  5a-\ . 

xy 


xy 


56 


FRACTIONS. 


Otherwise,  by  adding  the  integral  and  th e fractional  parts  sepa- 
rately.— There  will  be  less  liability  to  error  if  we  change  the  sign  be- 
fore the  fraction  in  the  second  quantity  to  +>  and  change  the  sign  of 
its  numerator ; thus 


+ — , (84). 

y 

i —c 

Then,  reducing  — and  — to  a common  denominator,  they  become 
x y 

by  , —cx 

— and  . 

xy  xy 

Adding  these  fractions  together,  and  adding  together  the  integral 
parts  2 a and  3a,  we  obtain 

ly—cx 

o a-j , as  beiore. 


Improper  fractions  in  the  results  obtained  by  this  Rule,  should  be 
reduced  to  integral  or  mixed  quantities  ; and  proper  fractions,  to  their 
lowest  terms. 


E XERCJSES. 


. . ,,  . 3a  2a  . 3a+2 

1.  A.dd  together  — , — , and  — - — 

T U O 


d 2 2 (jP* 

2.  Add  together  2x-\ , and  ix-\ — — 

4 o 


3.  Add  together  ?/2+  ^ , and  3 y2  — . 

O ut 


26-3 


4.  Add  together  — , 2a2,  and 

2x  5 


Ans.  2a + 


6a+8 

12 


Ans.  6x4- 


13a2 
20  ' 


Ans.  4 y2-\- 


5x 

12 


Ans.  2a2 + 


5a2  -\-Abx — 6x 
lQx  ' 


5.  Add  together  3x2  -j-  and  —j— 


Ans.  3 x2+a+  — . 

4 


FRACTIONS. 


57 


6.  Add  together  x2y,  — , and  x2y-\-  — . 

o « 


x a 2 — 3a: 

7 Add  together  2a2  — — and  — - — 


_ . ,,  a2  2>a  a2—  4 

8.  Add  together  — , — , and  - — - — . 

y y y2 


„ 2a2 + 3 

Ans.  2 x2y-\ — 


Ans.  2«2  + 


2<z2 — lx 


Ans 


a2y+3ay-\-a2  — 4 


o* 1 3d?-l“4 

0 Add  together  — - , 2,  and  — — 

O O 


2 a3  3 a3 

10.  Add  together  2 y-\ — — and  y — . 


3a;  a2-f-l 

ll.  Add  together  5a2  , — , and  — ^ — • 


. qq  2 /p  2 

12.  Add  together  10,  — — , and  3 — . 

3 o 


,,  , a2-\-x  , «2 — a:+l 

13.  Add  together  — - — and  - 

o 4 


o 34-w2 x2 

14.  Add  together  2y2--  and  — ^ 

CL  D 


r 


Ans.  2+ 


14z+7 

15 


alns.  3 y-  — . 


Ans.  5a2- 


3ca:+a2  + l 


8a:2 

Ans.  13- — 
15 


Ans. 


la2  — |—  a: — | — 3 
12  ~ 


. n . 15— 3a— ay2-\-ax2 

Ans.  2 y2 5a  ~ 


4 


58 


FRACTIONS. 


SUBTRACTION  OF  FRACTIONS. 

(9G.)  The  Difference  of  two  Fractions  is  found  by  means  of  a com 
mon  denominator. 

mi  a i t r & i b—a 

Thus  — subtracted  from  — leaves . 

x x x 

For  the  last  fraction,  - — — , added  to  — the  one  subtracted,  pro 
x x 

duces  — the  one  from  which  the  subtraction  is  made. 
x 

We  have  therefore 

RULE  XIV. 

(97.)  For  the  subtraction  of  Fractions. 


1.  If  the  fractions  have  not  a common  denominator , reduce  them 
to  a common  denominator. 

2.  Subtract  the  numerator  of  the  fraction  to  be  subtracted  from  the 
other  numerator,  and  place  the  Difference,  as  a numerator,  over  the 
common  denominator. 

3.  A mixed  quantity  may  be  taken  in  subtraction  under  the  form 
of  an  improper  fraction  ; or  the  integral  and  the  fractional  part  may 
be  taken  separately  in  subtracting. 


EXAMPLE. 

From  8a:  + — to  subtract  3a: — - — . 

V w 

Reducing  these  mixed  quantities  to  improper  fractions,  they  become 

8 xy-\-a  3 xw — b 

and . 

y w 

Reducing  these  fractions  to  a common  denominator, — subtracting 
the  second  of  the  resulting  numerators  from  the  first, — and  placing  the 
difference  over  the  common  denominator,  we  find 

5xywfaw-\-by  , „ aw-\-bu 

— , equal  to  5xf - . 

yw  yw 


FRACTIONS. 


59 


Otherwise,  by  taking  the  integral  and  the  fractional  parts  sepa- 
rately.— To  diminish  the  liability  to  error  in  adjusting  the  signs , we 

change  the  fraction  in  the  second  quantity  to  + — - , (84). 

Then,  reducing  — and to  a common  denominator,  they  become 

V w 

aw  , — by 

— and . 

yw  yw 

Subtracting  the  second  of  these  fractions  from  the  first,  and  3x  from 
,8a:,  we  find  the  difference  of  the  given  quantities  to  be 

5*+  a^±b!  , as  before. 
yw 

In  all  subsequent  exercises,  improper  fractions  in  the  results  should 
jbe  reduced  to  integral  or  mixed  quantities  ; and  proper  fractions,  to 
their  lowest  terms. 


EXERCISES. 


, „ a2  + 2 , h- 3 

1.  rrom  — - — subtract 

3 2 a 


„ „ 3a+x  , a— x 
2.  from  subtract  - — - 


3.  From subtract 

1+x  1 — x 


b c 

1.  From  3a  d subtract  2a . 

* y 


5.  From  2a:2 — subtract  a:2 — , 

* y 


Ans. 


2a3+4a  — 35  + 9 
6a 


Ans. 


3ay-\-xy—a+x 

y 2 


Ans. 


a 2 — a2x—b—bx 
1 —x2 


Ans.  a+ 


by+cx 


xy 


Ans.  x2  ■ 


ay—bx 

xy 


60 


FRACTIONS. 


6.  From  — subtract  -5 5 . 

x—y  x 2 — y 2 


From  3 y2 — — subtract  2 y2+—  . 

y * 


S.  From  2 a-\-i+  ^ subtract  3b  — ^ . 


9.  From  4a;2 — y — — subtract  y-\-  — 

O £ 


Ans. 


x-}-y—l 
x2  — y 2 


Ans.  y2  — 


ax+by 

xy 


Ans.  2a— 25+ 


X+V 


Ans  4a2— 2 y— 


2a3+3c3 


10.  From  252  + subtract  l2  — — 

4 O 


11.  From  5a3-) subtract  3a3 -| — 

y+i  2/— 1 


I 1 2 

12.  From  3a2 — — subtract  — a2+  ~ , 

o 


alrcs.  52+^?_i 


2a3 


2x 


Ans.  4x2  — 


y2  — 1 

3<z+35+2a2 


13.  From  253 — subtract  b2 , 

3 4 


Ans.  2b3— b2  — 


1 + 7 a 
12  ' 


(]  2 _J_  iyi  2 /jr2  ^2 

14.  From  5-\ subtract  l-\ . 

y y2 

Ans.  4+ 


a2y-\-x2y — a2 +a2 
y7 


FRACTIONS. 


61 


MULTIPLICATION  OF  FRACTIONS. 

(98.)  The  Product  of  two  or  more  Fractions  is  equal  to  the  product 
of  their  numerators  divided  by  the  product  of  their  denominators 


m,  a b ah 

Thus  multiplied  by  — produces  . 

xz  y x2y6 


For  the  first  of  these  fractions  is  equivalent  to  ax  2,  and  the  second 
to  by~3,  (77) ; and  the  product  of  these  tivo  equivalents  is 

abx~2y~3,  (40). 

And  by  transferring  the  factors  x~2  and  y"3  to  the  denominator, 
we  have 

~~  , ab  divided  by  x2y3,  (78). 
x2y3 

From  the  preceding  it  follows,  that 

(99.)  Multiplying  by  a Fraction  finds  such  a part  of  the  multipli- 
cand as  is  expressed  by  the  multiplier. 

Thus  a X A or  — X A produces  oi  a. 

i /w 


Hence 


Compound  Fractions. 


(100.)  A Fraction  multiplied  by  a fraction,  or  divided  by  an  inte- 
ger, may  he  expressed  by  a Compound  fraction,  that  is,  a fraction  of 
a fraction. 


For  example,  — x|  is  £ of  — , (99);  also  — -~2  is  A of  — . 

X X x * x 


(101.)  Multiplying  two  or  more  Fractions  together  is  equivalent  to 
reducing  a compound  to  a simple  fraction. 


Thus  — x£is£of  — , equal  to  — 
* ® ^ 2x 


From  proposition  (98),  we  have  the  following  Rule. 


62 


FRACTIONS. 


RULE  XV 

(102. ) For  the  Multiplication  of  Fractions. 


1.  Multiply  the  numerators  together  for  a numerator,  and  the  de- 
nominators together  for  a denominator. 

2.  An  integral  quantity  and  a fraction  are  multiplied  together,  by 
multiplying  the  numerator , or  dividing  the  denominator,  by  the 
integer. 

3.  A mixed  quantity  may  he  taken  in  multiplication  under  the 
form  of  an  improper  fraction  ; or  the  integral  and  the  fractional  part 
may  be  taken  separately  in  multiplying. 

It  may  also  be  remarked  that 

(103.)  A Fraction  is  multiplied  by  its  own  denominator , by  merely 
canceling  the  denominator. — And  equal  factors  may  be  canceled  in  a 
numerator  and  its  own  or  the  other  denominator,  without  altering  the 
Product  of  the  two  fractions,  (81). 

EXAMPLE. 


To  multiply 


by 


2 V 

a2  -\-2ax-\-x2 


The  first  numerator  is  equal  to  ( a-\-x)(a — x),  (58), 

and  the  second  denominator,  to  (a-{-x)  (a+x),  (59), 

By  canceling  the  factor  a-\-x  from  these  terms,  and  the  factor  y 
from  the  other  numerator  and  denominator,  the  operation  is  reduced  to 


a—x  2 2a  — 2x 

— — - X ; winch  produces  . 

3 y a+z  3ay+'3xy 


EXERCISES. 

3 

1.  Multiply  together  a2 — x2  and . 

a—x 

2.  Multiply  together  a2y  and  2+f-. 


Ans.  3a-f-3i. 
Am.  A„+  | 


3.  Multiply  together  and . 

3 a — x 

4.  Multiply  together  5y2  and  y2  — ~ . 

o 


. 2a  + 2x 

Ans.  — • 

oa — ox 

5ay2 

Ans.  5y* 0 — ]• 


FRACTIONS. 


63 


a-\-b  a — b 

b Multiply  together  — — ana  -7 

OX  • CL  "I”  0 

6 Multiply  together  2a2 — 3 a:  and 


3 b 


y~ 1 

7.  Multiply  together  a- 1 and  x~ 


S.  Multiply  together 


a — x 
ai-bi 


a-\-x 


ci-\~b 


and 


ab-b2 


9.  Multiply  together  5y 2 and  by2  -| - . 

X X 

&X  *7  CL 

10.  Multiply  together — and 


14 


2a:3 — 3a: 


Ans. 


a—b 


Ans. 


3a:2 
6a2b  — 9bx 


Ans. 


y — 1 

72  -yi2 


Ans. 


a * — x * 
a*+a2b2 


a * 
x 4 

3ax  — 5a 


Ans.  2 5yi 

Ans. 


4a:2— 6 


DIVISION  OF  FRACTIONS. 

(104.)  The  (Quotient  of  two  Fractions  is  equal  to  the  dividend  mul- 
tiplied by  the  reciprocal  of  the  divisor,  (80). 

CL  X CL  V 

Thus  the  Quotient  of  -7-  divided  by  — is  equal  to  — X — . 

b V b x 

For  since  the  value  of  the  dividend  is  not  altered  by  multiplying 
each  of  its  terms  by  both  terms  of  the  divisor,  (81),  the  quotient  is 
equal  to 

axy  x 
bxy  ' y ‘ 

And  by  dividing  the  numerator  of  this  dividend  by  the  numeratoi 
of  the  divisor,  and  the  denominator  by  the  denominator,  we  have  the 

ay  . . a y 

quotient  -f~- , which  is  equal  to  — X — . 

OX  0 cc 

Observe  that  the  Quotient  multiplied  into  the  divisor,  produces 
the  dividend,  (46.) 

ay  x axil  , a 

Ihus  — X — produces  7 — , equal  to  . 
bx  y bxy  b 


64 


FRACTIONS. 


Complex  or  Mixed  Fractions. 

(105.)  When  the  dividend  or  the  divisor  is  a Fraction  or  a mixed 
quantity,  the  dividend  over  the  divisor,  with  a line  between  them, 
forms  a complex  or  mixed  fraction. 

a 

Thus  4-  -i-(zFv)  equals , numerator  %■ , denominator  x+y. 

b ^ x-\-y  o 

From  the  nature  of  Division,  and  the  proposition  before  demonstra- 
•,ed,  (104),  we  have 

RULE  XVI. 

(10G.)  For  the  Division  of  Fractions. 

1.  Divide  the  numerator  of  the  divisor  into  the  numerator  of  the 
dividend,  and  the  denominator  into  the  denominator ; or  multiply  the  i 
dividend  by  the  reciprocal  of  the  divisor. 

2.  A fraction  is  divided  by  an  integral  quantity,  by  dividing  the  J 
numerator,  or  multiplying  the  denominator,  by  the  integer. 

3.  An  integral  quantity  is  divided  by  a fraction,  by  dividing  the 
integer  by  the  numerator,  and  multiplying  by  the  denominator  ; or  by  | 
multiplying  the  integer  by  the  reciprocal  of  the  fraction. 

4.  A mixed  quantity  may  be  taken  in  division  under  the  form  of  1 
an  improper  fraction : or  the  integral  and  the  fractional  part  may  be  { 
divided  separately 


To  divide 


10c2 


EXAMPLE. 

5c3 


a 2 — 2ax-\-x 2 


by 


2 -a2 ' 


The  dividend  must  be  multiplied  by 


a-  — 


5c3 


, which  is  the  recipro 


lal  of  the  divisor. 


If  we  cancel  the  factor  a — a;  from  a2 — 2 axf-x2,  (60),  and  from 
a2 — x2,  (58);  and  also  cancel  the  factor  5c2,  the  operation  will 
be,  (103), 

2 a+z  , . , , 2a+2x 

X ; which  produces  . 

a — x c ac — cx 

By  thus  canceling  common  factors,  vre  find  the  Quotient  in  its 
lowest  terms. 


FRACTIONS. 


65 


1.  Divide  2a2  — 2y2  by 


EXERCISES. 

g+y 

5cx 


_ 3a  3 

2 Divide  — t by 

a2—x 2 a—x 


bx—a 

3.  Divide  by  — - — 

a b 


a2-\-2abJrb2  a-\-b 

4-  Dm4e  — — by  tv 


_ 6a 2b  , 3a2 

J).  Divide  — ; — - by 


x2  — 2xy-\-y2  x— y 


6 Divide  3a2+^-by  a2-  ^ 
2 o 


7,  Divide  -i?L  by  * 


a3  + #3 


„ , 5a2y—\0ay2 + 5a  5a 

8.  Divide f- by  —= 

ax 2 x* 


n ^ —I—  T)  3 

9.  Divide  - — — - hyabfb2. 
2-\-3x 


10.  Divide  2a2 -\-\ay-\-2y2  by 


a+y 


Ans.  \Qacx—\Qcxy 


Ans. 


Ans. 


a-\-x 

abx-fb2 
abx — a2 

a+5 


Ans. 


Ans. 


2 ax 


2b 

x—y 


Ans.  3- 


Ans. 


Ans.  y— 


9.t2  + 3c 
6a2  — 2x2 

2x 

x2  —ax-\-a2 
2t/2-1 


Ans. 


a2  — ab-\-b 2 
26+3&r 


Ans.  \a-\-\y 


66 


FRACTIONS. 


11.  Add  ^ t°  - , and  divide  the  Sum  by 

12.  Add  a-  J to  and  divide  the  Sum  by  — . 

5 4 3 10 


Ans. 


4a+3ot 


Ans 


9 y 

9 a-\-y 
2x2 


c2  2x2  — 1 


at4  — 4 


13.  Add  — —to  — — , and  divide  the  Sum  by  — - 

2 o 2 


Ans. 


3(^  + 2)- 


, , . ..  ax  2ax-\-\  (ax-\- 2)2 

14.  Add  — — to  — , and  divide  the  Sum  by . 

4 J O 


Ans.  2(ux+2) 

y rty  2x 

15.  Subtract  4 from  — , and  divide  the  Remainder  by  — . „ 

3 2 J 3 3ay—2y. 

Ans.  4X 

y 2 ^2  — |—  y 

16.  Subtract  — from  — , and  divide  the  Remainder  by . 

4 4 3 2 

Ans. 


X*  CL“ 

17.  Subtract  — from  — , and  divide  the  Remainder  by 


a-\-z 


Ans. 


3 {a—x) 


_ , 2 ab — 3 ,5a  ,5 

18.  Multiply by  — and  divide  the  Product  by  — . 

* x ' x 2a2b-*-3a 

A, is.  - 

19.  Multiply  a~-  by  -4-,  and  divide  the  Product  by  . 

y2  x2  3 


Ans. 


20.  Multiply  by  - , and  divide  the  Product  by  A. 


a + b 


O 1 

— v ‘ 


a-  — x 


Ans. 


9-i  9 — ?/^  3— ?/ 

91.  Multiply  — - — by  — — , and  divide  the  Product  by 


% 


Ans.  27  -\-9y-\-3y2  -|-y3. 


67 


CHAPTER  V. 

SIMPLE  EQUATIONS. 


(107.)  An  Equation  is  an  expression  denoting  the  equality  of  two 
quantities  by  means  of  the  sign  =,  equal  to,  placed  between  them. 

The  quantity  on  the  left  of  the  sign  = is  called  the  first  member , 
or  side,  and  that  on  the  right  the  second  member,  or  side,  of  the 
Equation. 

Thus  3x+a5=:5a;+8c? — 9 is  an  equation, 
in  which  3x -\-ab  is  the  first  member,  and  oar+8 d — 9 is  the  second. 


(108.)  Equations  are  employed  in  the  solution  of  particular  mathe- 
matical questions,  or  in  the  investigation  of  general  mathematical 
principles. 

In  the  solution  of  questions,  the  unknown  or  required  quantity  is 
represented  by  a letter,  usually  x,  or  y,  Ac.,  and  an  Equation  is  then 
formed  which  expresses  the  relation  between  this  and  the  known  or 
given  quantities. 

To  give  a simple  example; — Suppose  we  wish  to  find  a number 
the  third  and  fourth  of  which  shall  together  make  35. 

. Let  x represent  the  number  to  be  found,  and  the  Equation 
, x x 

will  be  — + — = 35. 

O “E 


(109.)  The  solution  of  an  Equation  consists  in  finding  the  value  of 
the  unktiown  quantity  in  the  equation. 

The  value  found  for  the  unknown  quantity  is  verified , or  the  Equa- 
tion satisfied,  when  this  value,  substituted  for  its  symbol  in  the  equa- 
tion, makes  the  first  member  the  same  as  the  second. 

The  value  of  x in  the  preceding  equation  is  60,  since  this  number, 
substituted  for  x,  satisfies  the  equation ; thus 


60  60 

T + T =35- 

The  mode  of  solution  will  vary  with  the 


Different  Degrees  of  Equations. 

(110.)  A simple  Equation,  or  an  equation  of  the  first  degree,  is  one 
which  contains  no  power  of  the  unknown  quantity  but  its  first  pou-er. 

D+ 


68 


SIMPLE  EQUATIONS. 


3x-]-ax — 4 = 20  is  a simple  equation. 

A quadratic  Equation,  or  an  equation  of  the  second  degree,  is  one 
in  which  the  highest  power  of  the  unknown  quantity  is  its  second 
power  or  square. 

2a:2 -f-3aa:+5  = 30  is  a quadratic  equation. 

A cubic  Equation,  or  an  equation  of  the  third  degree,  is  one  in 
which  the  highest  power  of  the  unknown  quantity  is  its  third  power, 
or  cube;  and  so  on. 

Equations  are  also  distinguished  as 

Numerical , Literal , and  Identical  Equations. 

(111.)  A numerical  Equation  is  one  in  which  all  the  know?i  quan- 
tities are  expressed  by  numbers. 

2x-{-5x=25  — 3 is  a numerical  equation. 

A literal  Equation  is  one  in  which  some  or  all  of  the  known  quan 
tities  are  represented  by  letters. 

2x-\-ax=25  — 35  is  a literal  equation, 
in  which  a and  b are  supposed  to  represent  quantities  whose  values 
are  known. 

An  identical  Equation  is  one  in  which  the  two  members  are  the 
same,  or  become  the  same  by  performing  the  operations  which  are  in- 
dicated in  them. 

Thus  3a:  — 3ab  = o(x  — ab)  is  an  identical  equation. 

Transformation  of  Equations. 

(112.)  The  transformation  of  an  Equation  consists  in  changing  its 
form,  without  destroying  the  equality  of  the  two  members, — for  the 
purpose  of  finding  the  value  of  the  unknown  quantity,  or  of  discovering 
some  general  truth  or  principle. 

These  transformations  depend,  for  the  most  part,  on  the  following 

Axioms. 

(113.)  An  Axiom  is  a truth  which  is  self-evident, — neither  admit- 
ting nor  requiring  any  demonstration ; such  as, 

1 . Things  which  are  equal  to  the  same  thing,  are  equal  to  each 
other. 

2.  If  equals  be  added  to  equals,  the  sums  will  be  equal. 

3.  If  equals  be  taken  from  equals,  the  remainders  will  be  equal. 

4.  If  equals  be  multiplied  by  equals,  the  products  will  be  equal . 

5 If  equals  be  divided  by  equals,  the  quotients  will  be  equal . 

6 Any  like  powers  or  roots  of  equal  quantities,  are  equal 


SIMPLE  EQUATIONS. 


69 


Solution  of  Simple  Equations  Containing  but  one 
Unknown  Quantity. 

(114.)  The  value  of  the  unknown  quantity  is  found  by  making  its 
symbol  stand  alone  on  one  side  of  tbe  Equation,  so  as  to  be  equal  to 
known  quantities  on  tbe  other  side. 

In  order  to  this,  the  following  transformations  may  he  necessary , 
or  at  least  may  be  expedient. 

1.  Clearing  the  Equation  of  Fractions. 

2.  The  Transposition  and  Addition  of  Terms. 

3.  Changing  the  Signs  of  all  the  Terms  in  the  Equation. 

4.  Dividing  the  Equation  by  the  Coefficient  of  the  Unknown 
quantity. 

We  shall  apply  each  of  these  transformations  to  the  solution  of  the 
same  Equation. 


Clearing  an  Equation  of  Fractions. 


(115.)  An  Equation  is  cleared  of  fractions  by  multiplying  each  nu- 
merator into  all  the  denominators  except  its  oivn — regarding  each  inte- 
gral term  as  a numerator, — and  omitting  the  given  denominators 
Let  the  Equation  he 


x 

"3 


Multiplying  the  numerator  of  each  fraction  hy  the  denominators  of 
the  other  two,  and  the  integral  terms  7 and  x by  all  the  denominators, 
we  obtain 


27x  — 252  = 36*  — 335+  12a;. 

The  equality  of  the  two  members  is  not  destroyed  in  thus  clearing 
the  Equation  of  fractions,  because  each  of  the  terms  connected  by  the 
signs  + and  — in  the  two  members,  is  thus  multiplied  by  all  the 
denominators.  (103)  (113. ..4). 


An  Equation  may  also  be  cleared  of  fractions  by  multiplying  its 
two  members  by  the  least  common  multiple  of  the  denominators ; — 
observing  that  a fractional  term  will  be  multiplied  by  multiplying  its 
numerator  into  the  quotient  of  said  multiple  -y  the  denominator. 


In  tbe  given  Equation  the  least  common  multiple  of  the  denomina- 
tors is  12.  Multiplying  hy  12.  we  find 

9a:  — 84  = 1 2x  — 1 1 2 + 4x. 

The.  advantage  of  this  method  is.  that  the  new  equation  is  found 
in  its  loivest  terms. 


70 


SIMPLE  EQUATIONS. 


Transposition  and  Addition  of  Terms. 

(116.)  Any  term  may  be  transposed  from  one  side  of  an  Equation 
to  the  olher  by  changing  its  sign. — All  the  similar  terms  may  thus  be 
placed  on  the  same  side,  and  then  added  together. 

In  the  last  Equation  9a;  — 84=  12a; — 112  — |- 4ar,  by  transposing  — 84 
to  the  second  member,  and  12a:  and  4a:  to  the  first,  we  have 
9a: — 12a:  — 4a:  = — 112  + 84  ; 

And  by  adding  together  the  similar  terms, 

— 7a:  = — 28. 

The  equality  of  the  two  members  is  not  destroyed  by  transposing  a 
term  with  its  sign  changed  from  one  side  to  the  other,  because  this  is 
equivalent  to  adding  the  term  with  its  sign  changed  to  both  sides. 

Thus  by  adding  84  to  both  members  of  the  equation 
9a:— 84  = 12*— 112+4®, 

the  term  — 84  is  canceled  in  the  first  member  (28).  In  like  manner 
by  adding  — 12a:  to  both  members,  12a;  is  canceled  in  the  second  mem- 
ber ; so  also  with  4r.  (113. ..2). 

From  the  preceding  principles  it  follows,  that 
Tivo  equal  terms  with  like  signs  on  opposite  sides  of  the  sign  =,  may 
be  at  once  suppressed  from  the  Equation. 

Change  of  the  Signs  in  an  Equation. 

(117.)  All  the  signs  in  an  Equation  may  be  changed,  + to  — and 
— to  +,  without  affecting  the  equality  of  its  two  members. 

This  follows  from  the  principle  of  Transposition,  (116),  since  in 
transposing  all  the  terms , the  signs  would  all  be  changed,  but  the  two 
members  would  still  be  equal 

In  the  Equation  already  found 

— 7a:=-28, 

we  shall  have,  by  Transposition, 

28  = 70'-,  or  7a:=28. 

The  only  Transformation  which  remains  towards  finding  the  value 
of  x in  the  equation  at  first  assumed,  is  that  of  dividing  by  the  coeffi- 
cient of  2,  the  unknown  quantity.  (113. ..5). 

Dividing  both  members  of  the  preceding  Equation  by  the  coefficient 
of  x,  we  lind 

— 28  28 

x—  - — — =4  ; or  x=  — =4. 

— 7 7 

We  have  thus  found  the  value  of  x to  be  4.  This  value  may  bo 
verified  by  substituting  it  for  x in  the  original  equation. 


SIMPLE  EQUATIONS. 


71 


We  may  now  give 

RULE  XVII. 

(118.)  For  the  Solution  of  a Simple  Equation  containing  but  one 
unknown  quantity. 

1.  Clear 'the  Equation  of  fractions,  if  it  contains  any. 

2.  Transpose  all  the  terms  containing  the  unknoivn  quantity  to 
one  side,  and  all  the  known  terms  to  the  other  side,  of  the  equation. 

3.  Add  together  all  the  similar  terms  in  each  member. 

4.  Divide  both  members  by  the  coefficient  of  the  unknown  quanti- 
ty ; — observing  that  when  the  unknown  quantity  is  found  in  two  or 
more  dissimilar  terms,  its  coefficient  will  be  the  sum  of  its  coefficients 
in  those  terms. 

Note. — When  the  sum  of  the  terms  containing  the  unknown  quantity, 
after  transposition,  is  negative,  it  will  generally  he  expedient,  though  it  is 
never  necessary,  to  make  it  positive  by  changing  all  the  signs  in  the 
equation. 

EXAMPLE. 


Given 


x— 3 
2 


+ T=“- 


x+19 

2 ’ 


to  find  the  value  of  x. 


Clearing  the  equation  of  fractions,  by  multiplying  it  by  the  least 
common  multiple  of  the  denominator,  which  is  6,  we  have 
3x— -9  + 2x  = 120 — 3x— 57. 

By  transposition, 

3x+2x  + 3x— 120—  57  + 9. 

Adding  similar  terms, 

8x  = 72. 

Dividing  by  the  coefficient  of  x, 


Remark. — The  student  is  apt  to  err  in  Clearing  an  Equation  of  its  frac- 
tions, when,  as  in  this  Example,  a fraction  preceded  by  the  negative  sign 
has  a polynomial  numerator. 

The  sign  — before  the  fraction  in  the  second  member  above,  de 
notes  that  the  fraction  is  to  be  subtracted.  When  this  fraction  is  mul- 
tiplied by  the  6,  the  product  3x-\-51  is  subtracted  by  changing  its 
signs.  This  gives  the  terms  — 3x— 57  in  the  new  equation. 


12 


SIMPLE  EQUATIONS. 


EXERCISES. 

Numerical  Equations. 

1.  (riven  4x  — 8 = 13  — 3#  to  find  the  value  of  2. 

2.  Given  7£+l7  = 10£ — 19  to  find  the  value  of  x. 

3.  Given  8£+6  = 36 — 7x  to  find  the  value  of  x. 

4.  Given  59  — 7£=4£+26  to  find  the  value  of  a;. 

5.  Given  20 — 4x — 12  = 92  — 10#  to  find  the  value  of  x. 

Ans.  x=14 

6.  Given  8 — 3£+ 12  = 30 — 5£+4  to  find  the  value  of  x. 

Ans.  £=7 


Ans.  £=3 
Ans.  £=12. 
Ans.  £=2. 
AnS.  £ = 3 


7.  Given  — + 24=^  to  find  the  value  of  x. 
4 2 


8.  Given  -^-  + 4 =13 to  find  the  value  of  x. 

£ o 4 


££ £j  284 jy 

9.  Given b6.r  = to  find  the  value  of  x. 

4 5 


Ans.  £=194. 

O 

Ans.  £=1-2. 


Ans.  £=9 


10.  Given  + — ~~  =16 to  find  the  value  of  x. 

2 o 4 


Ans.  £=13. 


5 x 

11.  Given  3* — — =£+  - + 13-J  to  find  the  value  of  x. 

o O 


rl;w.  £=10. 


12.  Given  — 2£=15 to  the  value  of  x. 

O b 


2 Q $ W x 

13.  Given  — — — =£d ——  to  find  the  value  of  x. 

& • O 

f)/y> — in  ip 4-fj* 

14.  Given  ■ — = — |-£  to  find  the  value  of  x 


J./«S.  £=12 


A)is.  £=5 


.r  = 4.  |i 


15.  Given  -X- ^ --  = - — — to  find  the  value  of  r. 


A.  cS.  J.  ~ -i 


SIMPLE  EQUATIONS. 


7? 


16.  Given  — — 16=  — i — - to  find  the  value  of  x. 

3 5 4 


L7  Given  — — — = x — 3 — to  find  the  value  of  x. 

7 3 

18.  Given  — =1+  - to  find  the  value  of  x. 

3 2 

19.  Given  x — — =3* — 1^  to  find  the  value  of  x. 


Ans.  £=41 


Ans.  x—8. 


Ans.  x=7. 


20.  Given  — — +5=  to  find  the  value  of  x. 

id  O O 


Ans.  x= 


Ans.  x—6. 


op  2 

21.  Given  — + 6£  = — — - to  find  the  value  of  x. 
3 5 


Ans.  x=—£-3 


__  „ . 3x — 5 „ 2x — 4 „ , . . „ 

22.  Given  x — =12 — to  find  the  value  oi  x. 

2 3 

Ans.  £=65 

„„  r,-  21— 3£  2(2£+3)  „ 5.-C+1  „ . . 

23.  Given  ' =6 to  find  the  value  of  x. 

3 9 4 

Ans.  £=3. 


An  Equation  in  which  the  unknown  quantity  is  found  in  every 
term,  with  different  exponents  in  different  terms,  may  often  he  reduced 
to  a simple  Equation  by  dividing  it  by  some  power  of  the  unknown 
quantity.  (113...5). 

Thus  if  2a3  = 10a;2, — 

by  dividing  by  a;2  we  have  2a;=10  ; hence  £ = 5. 


Ans.  x—5. 


x2  3a;2 

24.  Given  -f  3a;  = 7£ — _1 — to  find  the  value  of  x. 

5 5 

•m . 

25.  Given  ^ | 1„  fi„d  the  value  of  2. 

Ans.  £=  10-^,. 

26.  Given  — -f = —5 — — — to  find  the  value  of  z. 

3x  2x  x2  — 1 U£ 


74 


SIMPLE  EQUATIONS. 


Literal  Equations. 


27.  Given  ax  — c— 


x—b 

a-\-c 


to  find  the  value  of  x. 


Clearing  the  equation  of  fractions,  by  multiplying  it  by  the  deno- 
minator a-\-c,  we  have 

a2x — ac+acx—c1  =x  — b. 

By  transposition 

a2x-\-acx  — x=ac-\-c2  — b. 

Dividing  by  the  coefficient  of  x, 

ac+c2  — b 

x—  ~2~, 7* 

az  +ac—  1 

dx  cix 

28.  Given  x-\ —b to  find  the  value  of  x. 

c c 


Ans.  x— 


be 


cib  1 

29.  Given  be ——d to  find  the  value  of  x. 

x x 


Ans.  x~ 


a-\-c~\~d 
ab—  1 


30.  Given  3a: — a=x — — — to  find  the  value  of  x. 


Ans.  X— 


„ 2>ax7  — 2bx-\-ax 

31.  Given  4 abx2—  to  find  the  value  of  x. 


be  Ad ' 
3 a-\-d 

~e+y 


32.  Given 


.4/is.  x— 

x(a—b)  ab  x _ . , , _ 

— — =a- 1 — to  find  the  value  of  x. 


Ans.  X— 


a — 2 u 
12ab-Za' 

3o(5  + 4) 
6(a— 6)-|-4 ' 


Remark. — In  an  identical  Equation  the  unknown  quantity  has  no 
determinate  value , since  any  quantity  whatever  may  be  substituted 
for  it,  and  the  equation  will  be  satisfied 
Thus  in  the  equation 

3a:  — 5 — ox — 5, 

the  two  members  will  be  equal  whatever  be  the  value  of  x.  (113.  .3) 


SIMPLE  EQUATIONS. 


75 


PROBLEMS 

In  Simple  Equations  of  one  unknown  Quantity. 

(119.)  A Problem  is  a question  proposed  for  solution  ; and  the  solu- 
tion of  a problem  by  Algebra  consists  va.  forming  an  Equation  which 
shall  express  the  conditions  of  the  problem,  and  then  solving  the 
equation. 

The  general  method  of  forming  the  Equation  of  a problem,  is,  to 
represent  a required  quantity  by  x,  or  y , &c.,  and  then  to  perform  or 
indicate  the  same  operations  that  would  be  necessary  to  verify  the 
value  of  x or  y , supposing  that  value  to  have  been  found. 

EXAMPLES  AND  EXERCISES. 

1.  What  number  is  that  to  the  double  of  which  if  13  be  added,  the 
sum  will  be  75  ? 

Let  x represent  the  required  number  ; 
then  2x  will  represent  twice  the  number  ; 
and,  by  the  conditions  of  the  problem,  the  equation  will  be 

2z+13  = 75. 

The  value  of  x in  this  equation  is  the  number  required. 

Ans.  31. 

2.  Find  a number  such  that  if  it  be  multiplied  by  5,  and  24  be 
subtracted  from  tbe  product,  the  remainder  will  be  36.  Ans.  12. 

3.  What  number  is  that  to  of  which  if  25  be  added,  the  sum  ob- 
tained will  be  equal  to  the  number  itself  minus  39  l Ans.  96. 

4.  Find  a number  such  that  if  \ of  it  be  subtracted  from  three 

times  the  number,  the  remainder  will  be  77.  Ans.  28. 

5.  Find  what  number  added  to  the  sum  of  one  halftone  third,  and 
one  fourth  of  itself  will  equal  4 added  to  twice  the  number. 

• Ans.  48. 

6.  Divide  the  number  165  into  two  such  parts  that  the  less  may 
be  equal  to  of  the  greater. 

Let  x represent  the  less  part ; 
then  165 — x will  represent  the  greater  ; 
and  the  equation  will  be 


165 — x 


Ans.  15  and  150. 

7.  Divide  the  number  100  into  two  such  parts  that  six  times  the 
less  may  be  equal  to  twice  the  greater.  Ans.  25  and  75. 


76 


SIMPLE  EQUATIONS. 


8.  It  is  required  to  divide  75  into  two  such  parts  that  3 times  the 

greater  may  exceed  7 times  the  less  by  15.  Ans.  21  and  54. 

9.  What  sum  of  money  is  that  to  which  if  $100  he  added,  -I  of  the 

amount  will  he  $400  ? Ans.  $500. 


10.  A prize  of  $100  is  to  he  divided  between  two  persons. — the 
share  of  the  first  being  £ of  that  of  the  other.  What  are  the  shares  ? 

Ans.  $43f ; $561. 

11.  A post  is  J of  its  length  in  the  mud,  of  it  in  the  water,  and 
15  feet  above  the  water.  What  is  the  length  of  the  post  ? 

Ans.  36  feet. 

12.  Find  a number  such  that  if  it  be  divided  by  12,  the  divisor 

dividend  and  quotient  together  shall  make  64.  Ans.  48. 

13.  In  a mixture  of  wine  and  cider,  ■§•  of  the  whole  •plus  25  gallons 

was  wine,  and  1 part  minus  5 gallons  was  cider.  What  was  the 
whole  number  of  gallons  in  the  mixture  ? Ans.  120. 

14.  After  a person  had  expended  $10  more  than  -g-  of  his  money,  \ 
he  had  $15  more  than of  it  remaining.  What  sum  had  he  at  first! 

Ans.  $150. 


15.  Divide  the  number  91  into  two  such  parts  that  if  the  greater 
be  divided  by  their  difference,  the  quotient  may  be  7. 

Ans.  49  and  42. 

16.  A and  B had  equal  sums  of  money ; the  first  paid  away  $25, 

and  the  second  $60,  when  it  appeared  that  A had  twice  as  much  left 
as  B.  What  sum  had  each  ? Ans.  $95. 

17.  After  paying  away  of  my  money,  and  then  -3  of  what  was 

left,  I had  $180.  What  sum  had  I at  first?  Ans.  $300. 

18.  A line  37  feet  in  length  is  to  be  divided  into  3 parts,  so  that 
the  first  may  be  3 feet  less  than  the  second,  and  the  second  5 more  1 
than  the  third  ; what  are  the  parts?  A?is.  12,  15,  and  10  feet. 

19.  A can  perform  a piece  of  work  in  12  days,  and  B can  perform  1 
the  same  in  15  days.  In  what  time  could  both  together  do  the  work  ? 

Let  x represent  the  number  of  days.  Then  since  A could  do  ^ of 
the  work,  and  B ^3  of  it,  in  1 day, 
cc 

— will  represent  the  part  of  the  work  A could  do  in  x days, 


15 


will  represent  the  part  of  the  work  B could  do  in  x days 


The  equation  is  ^ =1,  the  entire  work.  Ans.  6 | days. 


SIMPLE  EQUATIONS.  77 

20.  If  A could  mow  a certain  meadow  in  6 days,  B in  8 days,  and 
L *4  5 days,  in  what  time  could  the  three  together  do  it  1 

Ans.  2 Ay  days. 

21.  Out  of  a cask  of  wine,  which  had  leaked  away  a third  part, 

20  gallons  were  afterwards  drawn,  and  the  cask  was  then  found  to  be 
but  half  full ; how  much  did  it  hold  ? Ans.  120  gallons. 

22.  It  is  required  to  divide  $300  between  A,  B,  and  C,  so  that  A 

may  have  twice  as  much  as  B,  and  C as  much  as  the  other  two  to- 
gether. Ans.  A $100,  B $50,  C $1 50. 

23.  A gentleman  spends  of  his  yearly  income  in  board  and 

lodging,  of  the  remainder  in  clothes,  and  then  has  $20  left.  What 
is  the  amount  of  his  income  ? Ans.  $180. 

24.  A person  at  the  time  he  was  married,  was  3 times  as  old  as 
his  wife,  but  15  years  afterwards  he  was  only  twice  as  old.  "What 
were  their  ages  on  their  wedding  day  ? Ans.  45  and  15  years. 

25.  Two  persons,  A and  B,  lay  out  equal  sums  of  money  in  trade  ; 

the  first  gains  $126,  and  the  second  loses  $87,  and  A’s  money  is  now 
double  of  B’s;  what  did  each  lay  out  ? Ans.  $300. 

26.  A courier,  who  travels  60  miles  a day,  had  been  dispatched  5 

days,  when  a second  is  sent  to  overtake  him,  who  goes  75  miles  a day, 
in  what  time  will  he  overtake  him  ? Ans.  20  days. 

27.  An  island  is  60  miles  in  circumference,  and  two  persons  start 
together  to  travel  the  same  way  around  it:  A goes  15  miles  a day; 
and  B 20  ; in  what  time  would  the  two  come  together  again  1 

Ans.  12  days. 

28.  A man  and  his  wife  usually  drank  out  a cask  of  beer  in  12 
days,  but  when  the  man  was  from  home  it  lasted  the  woman  30  days ; 
how  many  days  would  the  man  alone  be  in  drinking  it  ? 

Let  x be  the  number  of  days  ; 

then  — is  the  part  that  he  would  drink  in  1 day ; 

and  since  the  woman  would  drink  ^ of  it  in  1 day,  the  equation  will 

be 

1 = — , the  part  both  would  drink  in  1 day. 

i 30  12  f J 

Ans.  20  days. 

29.  If  A and  B together  can  do  a piece  of  work  in  9 days,  and  A 

alone  could  do  it  in  15  days,  in  what  time  ought  B alone  to  accomplish 
the  work  1 Ans.  22^  days. 

30.  The  hour  and  the  minute  hand  of  a clock  or  watch  are  exactly 
together  at  12  o'clock  ; when  are  they  next  together  1 

Ans.  minutes  past  one 


78 


SIMPLE  EQUATIONS. 


31.  It  is  required  to  divide  $1000  between  A,  B,  and  C,  so  that 

A shall  have  half  as  much  as  B,  and  C half  as  much  as  A and  B to- 
gether. Ans.  $222|  ; $444f  ; S333f. 

32.  A person  being  asked  the  hour,  answered  that  the  time  past 
noon  was  § of  the  time  till  midnight ; what  was  the  hour  ? 

Ans.  48  min.  past  4. 

33.  It  is  required  to  divide  the  number  60  into  two  such  parts  that 

their  product  shall  be  equal  to  3 times  the  square  of  the  less ; what 
are  the  parts  ? Ans.  15  and  45. 

34.  How  much  wine  at  90  cents  a gallon,  and  how  much  at 

$1.50  a gallon,  will  be  required  to  form  a mixture  of  20  gallons  which 
shall  be  worth  $1.25  a gallon  1 A?is.  8f  gal. ; Ilf  gal. 

35.  A cistern  is  supplied  with  water  by  three  pipes  which  would 

severally  fill  it  in  4,  5,  and  6 hours.  In  what  time  would  three  pipes  : 
running  together  fill  the  cistern  ? Ans.  Iff  hours. 

36  If  $1000  he  divided  between  A,  B,  and  C,  so  that  B shall 
have  as  much  as  A and  half  as  much  more,  and  C as  much  as  B and 
half  as  much  more,  what  will  be  the  portion  of  each  ? 

Ans.  $210ff ; $315ff;  $4731f. 

37.  A person  has  a lease  for  99  years,  and  f of  the  time  which  has 

expired  on  it  is  equal  to  | of  that  which  remains.  Required  the  time 
which  remains  on  the  lease.  Ans.  45  years. 

38.  A merchant  bought  cloth  at  the  rate  of  $7  for  5 yards,  which 

he  sold  again  at  the  rate  of  $11  for  7 yards,  and  gained  $100.  How 
many  yards  were  thus  bought  and  sold  ? Ans.  583f  yards. 

39.  A and  B together  possess  the  sum  of  $9800  ; and  five-sixths  of 

the  sum  owned  by  A is  the  same  as  four-fifths  of  that  owned  by  B. 
What  is  the  sum  owned  by  each  ? Am.  $4S00  ; $5000. 

40.  The  assets  of  a bankrupt  amounting  to  $5600  are  to  be  divided 

among  his  creditors  A,  B,  and  C,  according  to  their  respective  claims,  i 
A’s  claim  is  \ of  B’s,  and  C's  is  f of  B s ; what  sum  must  each  of  the  i 
creditors  receive ? Ans.  $1292^;  $2584^-;  $1/^3^. 


SIMPLE  EQUATIONS. 


79 


Simple  Equations  Containing  Two  or  more  Unknown 
Quantities. 

(120.)  It  is  sometimes  necessary  to  employ  two  or  more  unknown 
quantities  in  the  solution  of  a Problem ; and  in  this  case  there  must  be 
formed  as  many  independent  Equations  as  there  are  unknown  quanti- 
ties employed. 

Two  equations  are  said  to  he  independent  of  each  other  when  they 
express  essentially  different  conditions , so  that  one  of  the  equations  is 
not  a mere  transformation  of  the  other. 

Equations  which  thus  express  different  conditions  of  the  same  Prob- 
lem, are  sometimes  called  simultaneous  equations. 

Solution  of  Two  Simple  Equations  Containing  Two 

IUnknown  Quantities. 

(121.)  From  two  Equations  containing  two  unknown  quantities  we 
may  derive  a new  equation  from  which  one  of  those  quantities  shall  be 
eliminated,  or  made  to  disappear.  The  value  of  the  remaining  un- 
known quantity  may  be  found  from  the  new  equation-;  and  this  value 

I put  for  its  symbol  in  one  of  the  given  equations,  will  determine  the 
other  unknown  quantity. 

Elimination  by  Addition  or  Subtraction. 

. 

(122.)  The  two  terms  which  contain  the  same  letter  in  two  Equa- 
tions, may  be  made  equal  by  multiplying  or  dividing  the  equations  by 
proper  quantities.  That  letter  will  then  be  eliminated  in  the  sum , or 
tlse  in  the  difference,  of  the  new  equations. 

.EXAMPLE. 

Given  the  equations  2x  + 3y=23 
and  5x— 2y=  10, 
to  find  the  values  of  x and  y. 

Multiplying  the  first  equation  by  2,  and  the  second  by  3,  we  have 

4a;  + 6y=46, 

and  15a; — 6^/=30.  (113. ...4). 

Adding  together  the  corresponding  sides  of  these  equations,  we  find 

193=76.  (113. ...2). 

which  gives  X—4. 

Putting  4 for  x in  the  first  of  the  two  given  equations,  we  obtain 

8 + 3y= 23, 
which  gives  iy  = 5. 


80 


SIMPLE  EQUATIONS. 


Observe  that  if  6y  had  the  same  sign  in  the  two  equations  which 
were  added  together , this  term  would  have  been  eliminated  by  taking 
the  difference,  not  the  sum,  of  these  equations. 


Elimination  by  Substitution. 

(123.)  The  value  of  one  of  the  unknown  quantities  in  an  Equation, 
may  be  found  in  terms  of  all  the  other  quantities  in  the  equation.  If 
this  value  be  then  substituted  for  its  representative  letter  in  another 
equation,  that  letter  will  be  eliminated. 

EXAMPLE. 

Given,  as  before,  2x+3y=23, 
and  5x  — 

to  find  the  values  of  x and  y. 


We  will  find  the  value  of  x in  the  first  equation,  as  if  the  value  of 
y were  known. 

By  transposition,  2x—2o — 3 y ; 
dividing  both  members  by  the  coefficient  of  x, 

23  — 3 y 


We  now  substitute  this  value  of  x for  x in  the  second  equa- 
tion. In  doing  this,  we  must  multiply  this  fraction  by  the  coeffi- 
cient 5 in  the  first  term  of  that  equation. 

115  — \5y 

; —2y—10. 


Then 


find 


Clearing  this  equation  of  fractions,  transposing,  See.,  we  shall 


y=5. 

Putting  5 for  y in  the  first  equation, 

2z+15=23; 
which  gives  x=4. 

The  values  of  x and  y are  thus  found  to  be  the  same  as  before. 


This  method  of  Elimination  depends  on  the  evident  principle,  that 
equivalent  algebraic  expressions  may  be  taken,  the  one  for  the  other ; 
that  is,  equal  quantities  may  be  substituted  for  each  other. 


SIMPLE  EQUATIONS. 


81 


Elimination  by  Comparison. 

(124.)  If  the  value  of  the  same  letter  he  found  in  each  of  two 
Equations,  in  terms  of  the  other  quantities  in  the  equations,  that  letter 
will  be  eliminated  by  putting  one  of  these  values  equal  to  the  other. 

EXAMPLE. 

Given,  as  before,  2a:+3?/=:23, 
and  5x — 2?/=l0, 
to  find  the  values  of  x and  y. 

We  will  find  the  value  of  x in  each  equation,  as  if  the  value  of  y 
were  known. 

Transposing  and  dividing  in  the  first  equation, 

_ 23—3 y 
x-  — . 

Transposing  and  dividing  in  the  second  equation, 

10  + 2 y 

X~  5 

Putting  the  first  of  these  values  of  x equal  to  the  second, 

23—3?/  10  + 2?/  , „ 

which  will  give  y—5. 

By  substituting  5 for  y in  any  of  the  preceding  equations,  the 
value  of  x will  be  found  to  be  4,  as  in  the  two  preceding  solutions. 

Before  applying  any  of  the  preceding  methods  of  Elimination,  the 
Equations  should  generally  be  cleared  of  fractions,  if  they  contain  any  ; 
the  necessary  transpositions  must  be  made ; and  similar  terms  must 
be  added  together. 

Elimination  by  Addition  or  Subtraction  will  generally  be  found  the 
simplest  method,  since  it  is  free  fro m fractional  expressions,  which  are 
likely  to  occur  in  the  application  of  the  other  two  methods. 

Let  the  student  apply  each  of  the  three  methods  to  the  first  ten  ol 
the  following  Exercises. 

Equations  may  be  marked,  for  reference,  by  the  numbers  (1),  (2) 
(3),  &c.,  or  the  capitals  (A),  (B),  (C),  &c. 

Thus  (A)  2x+?/  = 10,  would  be  called  equation  (A). 

5 


82 


SIMPLE  EQUATIONS. 


EXERCISES. 

1.  Given  22+3?/  = 29,  and  2>x  — 2y=\\,  to  find  the  values  of  x 

and  y.  Ans.  2=7,  and  y— 5. 

2.  Given  ox — 3?/=9,  and  22+5?/=  1C,  to  find  the  values  of  i 

and  y.  Ans.  2 = 3,  and  y—2. 

3.  Given  x+2?/=l7,  and  32+?/=  16,  to  find  the  values  of  x 

and  y.  Ans.  2 = 3,  and  y = 7. 

4.  Given  42+?/  = 34,  and  10^  — 2=12,  to  find  the  values  of  x 

and  y.  Ans.  x = 8,  and  //  = 2. 

5.  Given  32+4?/=88,  and  62+5?/=  128,  to  find  the  values  of  x 

and?/.  Ans.  2=8,  and  2/=  16. 

6.  Given  72  + 3?/=42,  and  8 y — 22=50,  to  find  the  values  of  x 

and  y.  Ans.  2 = 3,  and  y—1. 

7.  Given  8 y — 32=29,  and  6 y — 42=20,  to  find  the  values  of  x 

and  y.  Ans.  x=l,  and  y= 4. 

8.  Given  62 — 5y  = 39,  and  lx  — 3?/=54,  to  find  the  values  of  x 

and  y.  Ans.  2 = 9,  and  y— 3. 

9.  Given  12x — 9z/  = 3,  and  122+ 16?/=228,  to  find  the  values  of  x 

and  y.  Ans.  2=7,  and  ?/= 9. 

10.  Given  52+7?/=201,  and  82 — 3y  = 137,  to  find  the  values  of  2 

and  y.  Ans.  2 = 22,  and  ?/=13. 

11.  Given  ly-\-  y =99,  and  72+  y =51,  to  find  the  values  of  2 

and  y.  A7is.  2 = 7,  and?/=14. 

12.  Given  =7,  and  — - + ^ =8,  to  find  the  values  of  2 


Ans.  2 = 6,  and  ?/=] 2. 

4?/  9 y 

13.  Given  64 — = — , and  77 — = — to  find  the  values  of 

do  6 10 


and  y. 

13. 

2 and  y. 

14.  G 
of  2 and  y. 


Ans.  2 = 60,  and  y=30. 

14.  Given  21  — 6?/=  , and  23—52=  ■ , to  find  the  values 

,4  3 

Ans.  2 = 4,  and  y=3. 

_.  32+4y  . 2 , 62—2?/  y 

15.  Given — - =10 , and  — - =14 — + , to  find  the 

5 4 3 6 


values  of  2 and  y. 

16.  Giv< 
of  2 and  y. 


A)is.  2=8,  and  y=i. 

_ . 22  3?/  22  3x  2y  7 , . , 

16.  Given 1 — - = — , and 1 — + =4  — . to  find  the  values 

3 5 5 5 3 10 

+;?s.  2=3,  and  y= 4. 

10—2  y— 10  . 2?/+4  22+?/  2+13 

'1,,d  = s^  = — 


17.  Given 


d — 2 


SIMPLE  EQUATIONS. 


olivll  LiJii  U A 1 IOjNS.  g<^ 

to  find  the  values  of  x and  y.  Ans.  x—1,  and  y=lO 

18.  Find  the  values  of  x and  y in  the  equations 

-m  , c c in  , 6a-— 35  80  + 3*  4x+3y-8 

5 lo  6 7 

Ans.  *=10,  and  y—15. 

19.  Find  the  values  of  * and  y in  the  equations 

y — a a — * 

x ; — —c,  and?/ — — — = d . 

b J b 

Clearing  the  equations  of  fractions,  we  have 
(A)  bx — y-\-a  = bc:  and  (B)  by — a-\-^~bd. 

By  transposition  in  these  two  equations,  we  find 
(C)  bx—y=bc — a,  and  (D)  bij-\-x=bd-\-a. 

Multiplying  equation  C by  b,  in  order  to  eliminate  y, 

(E)  b2x — by—b2c — ah. 

Adding  together  equations  D and  E,  (122.) 
b2x-\-x=b2c-\-bd — ai  + <z. 

Dividing  both  sides  of  this  equation  by  the  co-efficient  of  x, 
b2c-\-bd — ab-\-a 
, X=  b2  + 1 ' 

The  value  of  y will  be  found  if  we  multiply  equation  B by  b,  sub« 
:act  equation  A,  &c. 

b2d-\-ab— bc-tf-a 

y= ^TT— • 

x y 

20.  Given  — |-  - = 1,  and  bx-\-cy~de,  to  find  the  values  of  * and  ?/ 

a a 

ac — de  de—ab 

Ans.  x=  - — , and  ?/  = ■ — 

c-b  c—b 

. • A.b 

21.  Given  2c* — 4c  = — 3 dy,  and  lx— — , to  find  the  values  of  x 

a 

4 b 28ac  — 8bc 

* la  J 21  ad 

Q'lJ  Q YiV  d 

22.  Given  *-) = — , and  m =— . to  find  the  values  of  x and  y 

a a xx 

. nc+bd  me— ad 

Ans.  x = and  ?/=- 


na-\-mb 


na  -\-rnb 


E 


84 


SIMPLE  EQUATIONS. 


Solution  of  Three  or  more  Simple  Equations  containing 
as  many  Unknown  Quantities. 

(125.)  From  three  Equations  containing  three  unknown  quantities, 
we  may  derive  two  new  equations,  containing  but  two  of  the  unknown 
quantities — by  eliminating  one  of  the  unknown  quantities  from  the 
first  and  second  equations,  and  the  same  unknown  quantity  from  either 
of  these  and  the  third  equation. 

The  two  equations  thus  obtained  may  be  solved  as  already  exem 
plified,  (121).  Two  of  the  unknown  quantities  will  thus  be  deter- 
mined ; and  by  substituting  their  values  in  one  of  the  given  equations, 
the  value  of  the  other  unknown  quantity  may  be  readily  found. 

In  a similar  manner  four  equations  containing  four  unknown 
quantities,  may  be  reduced  to  three  equations  containing  but  three 
unknown  quantities ; and  these  three  may  then  be  reduced  to  two 
Five  equations  may  be  reduced  to  four , these  four  to  three,  and  these 
throe  to  two,  &c. 


example. 


(riven  the  equations  xf  y + z=  29, 
a:+2y+3z=  62, 

6a;  + 4i/+3z=r:120,  to  find  x,  y , and  z. 

Multiplying  the  first  equation  by  3,  in  order  to  eliminate  z, 

3 x -\-  3 y 3s  = 8 7 . 

Subtracting  this  equation  from  the  third  equation, 

3a?-|-y=33  5 

Subtracting  the  second  equation  from  the  third, 

5x+2y=58.  (122). 

The  last  two  equations  contain  but  two  unknown  quantities,  a:  and 
y,  from  which  we  may  find  a;  = 8,  and  y= 9. 

Substituting  these  values  of  x and  y in  the  first  equation, 

8Q9  + z = 29, 
which  gives  z=  12. 

If  the  value  of  x were  found  in  the  first  equation,  and  substitutec 
for  x in  the  second  and  third  equations,  we  should  then  have  twc 
new  equations  containing  y and  z,  (123). 

We  might  also  have  found  the  two  neio  equations  by  finding  the 
value  of  the  same  letter,  as  x,  in  each  of  the  three  given  equations 
then  putting  the  first  of  these  values  equal  to  the  second,  and  eitlie' 
the  first  or  the  second  equal  to  the  third,  (124) 


SIMPLE  EQUATIONS. 


85 


The  best  methods  of  elimination  to  he  adopted  in  particular  cases, 
can  he  learned  only  from  experience.  Regard  should  be  had  to  simpli- 
city and  brevity  in  the  operations  employed. 

EXERCISES. 

23.  Given  x-\-y-\-z~9,  a+2y-|-32  = 16,  and  x+2y-\-iz  = 2l,  to 

find  the  values  of  a;,  y,  and  z.  Ans.  a:  = 4,  y=  3,  2 = 2. 

24.  Given  x-\- y-\-z  — 18,  x+3?/-l-22  = 38,  and  a:+-|-;y-f-i|2:  = 10,  to 

find  the  values  of  x,  y,  and  2.  Ans.  a;=4,  ?/  = 6,  2 = 8. 

25.  Given  3*  — 9?/=33,  4y-|- 22  = 5.2—20,  and  11a;  — 7?/  = 37  + 62, 

to  hud  the  values  of  a:,  y,  and  2.  Ans.  x=2,  y=z — 3,  2=1. 

26.  Given  22  = 21 — ^(x+y),  3a;  = 72,  and  38=-J-(3a;+?/ — 2)  to  find 

the  values  of  x,  y , and  2 Ans.  £=24,  y — 9,  2 = 5. 


PROBLEMS 

In  Simple  Equations  of  one,  two,  8pc.,  Unknown 
Quantities 

(126.)  When  two  or  more  required  quantities  are  so  related  that, 
when  one  of  them  is  found,  the  others  may  be  conveniently  derived 
from  that  one,  the  Problem  may  be  most  readily  solved  by  a single 
Equation.  In  some  questions,  however,  it  is  necessary  to  represent  each 
of  the  required  quantities  by  an  appropriate  symbol,  and  then  to  form 
as  many  Equations  as  there  are  unknown  quantities  to  be  found. 

EXAMPLES  AND  EXERCISES. 


1.  Find  two  numbers  such  that  ^ of  the  first  with  ^ of  the  second 
shall  be  equal  to  9,  and  ^ of  the  first  with  | of  the  second  shall  be 
equal  to  5. 

Let  x represent  the  first,  and  y the  second  number,  then  by  the 
conditions  of  the  problem,  we  have 

x v , x y 

¥+3=9’  and  1 + 5 =5- 

The  values  of  x and  y in  these  equations  are  the  numbers  re- 
quired. . Ans.  8 and  15. 

2.  Divide  the  number  100  into  two  such  parts  that  ^ of  the  first 
and  ^ of  the  second  part  shall  together  make  30. 

Ans.  60  and  40 


86 


SIMPLE  EQUATIONS. 


3.  Find  two  numbers  such  that  their  sum  shall  he  60,  and  the  less 

number  4 of  the  greater.  Ans.  15  and  45. 

4.  At  a certain  election  946  men  voted  for  two  candidates,  and  the 

successful  one  had  a majority  of  558.  How  many  votes  were  given 
for  each  candidate?  Ans.  752  and  194. 

5.  Divide  the  number  48  into  two  such  parts  that  the  quotient  of 

the  greater  part  divided  by  4,  may  be  equal  to  4 times  the  quotiet  of  the 
greater  part  divided  by  the  less.  Ans.  32  and  16. 

6.  A,  B,  and  G make  a joint  contribution  which  in  the  whole  ( 

amounts  to  $400  ; B contributes  twice  as  much  as  A and  $20  more,  I 
and  C as  much  as  the  other  two  together.  What  sum  did  each  con- 
tribute ? Ans.  A $60,  B $140,  C $200. 

7.  Find  three  numbers  such  that  the  sunt  of  the  1st  and  2d  shall 

be  35,  the  sum  of  the  1st  and  3d  40,  and  the  sum  of  the  2d  and  3d 
45.  Ans.  15,  20,  and  25. 

8.  A sum  of  money  was  divided  between  A and  B,  so  that  B’s 
share  was  f of  A’s,  and  A’s  share  exceeded  f of  the  whole  sum  by 
$50.  What  was  the  share  of  each?  Ans.  As  $450,  B's  $270. 

9.  The  stock  of  three  traders  amounted  to  $760.  The  shares  of  , 
the  1st  and  2d  together  exceeded  the  share  of  the  3d  by  $240  ; and  i 
the  share  of  the  1st  was  $360  less  than  the  sum  of  the  shares  of  the  | 
other  two  ; what  was  the  share  of  each  ? 

A?is.  $200;  $300;  $260. 

10.  A man  being  asked  the  age  of  himself  and  son.  replied,  ■ If  I 

were  % as  old  as  I am  +3  times  the  age  of  my  son,  I should  be  45 ; I 
and  if  he  were  -j-  his  present  age  +3  times  mine,  he  would  be  111.'’ 
Required  their  ages.  Htts.  36  and  12. 

11.  A and  B together  have  $340,  B and  C together  $384,  and  A 
and  C together  $356  ; what  sum  has  each  ? 

Ans.  $156;  $1S4;  $200. 

12.  A number  which  is  expressed  by  two  digits  is  equal  to  4 times  4 
the  sum  of  its  digits,  and  if  1 8 be  added  to  the  number,  its  digits  will  i 
be  interchanged  with  each  other  ; what  is  the  number  ? 

Let  x represent  the  tens' , and  y the  units'  figure  of  the  number , 
then  10 x-\-y  will  represent  the  number. 

By  the  conditions  of  the  problem  the  equations  will  be 
10a:+?/  = 4(a:+y), 

10a:+^+l8  = 10y-fa:. 

Ans.  24. 


SIMPLE  EQUATIONS. 


87 


13.  It  is  required  to  divide  the  number  35  into  three  such  parts, 
that  £ of  the  first,  g-  of  the  second,  and  ^ of  the  third,  shall  all  be 
equal  to  each  other;  what  are  the  parts?  Ans.  8,  12,  and  16. 

1 4.  A and  B have  both  the  same  income  ; A saves  J of  his  an- 
nually, but  B,  by  spending;  $50  per  annum  more  than  A,  at  the  end  of 
4 years,  finds  himself  $100  in  debt ; what  is  their  income  ? 

Ans.  $125. 

15.  A gentleman  purchased  a chaise,  horse,  and  harness  for  $180  ; 
the  horse  cost  twice  as  much  as  the  harness,  and  the  chaise  twice  as 
much  as  the  horse  and  harness  together  ; what  was  the  price  of  each  ? 

Ans.  $120;  $40;  $20. 

16.  A farmer  purchased  100  acres  of  land  for  $2450  ; for  a part  of 
the  land  he  paid  $20  an  acre,  and  for  the  other  part  $30  an  acre. 
How  many  acres  were  there  in  each  part  ? 

Ans.  55,  and  45  acres. 

17.  What  fraction  is  that  to  the  numerator  of  which  if  1 he  added, 

the  value  will  be  \ ; but  if-  1 be  added  to  the  denominator,  the  value 
of  the  fraction  will  be  ^ ? Ans.  -g-. 

18.  A and  B together  possess  an  income  of  $570  ; if  A’s  income 
were  3 times,  and  B’s  5 times  as  much  as  each  really  is,  they  would 
together  have  $2350.  What  is  the  income  of  each  ? 

Ans.  $250  ; and  $320. 

19.  How  old  are  we?  said  a person  to  his  father ; 6 years  ago, 

leplied  the  latter,  I was  a third  more  than  3 times  as  old  as  you  were  ; 
but  in  3 years,  if  I multiply  your  age  by  21,  it  will  then  be  equal  to 
.mine.  What  were  their  ages  ? Ans.  15  and  36. 

20.  Find  a number  such  that  if  we  subtract  it  from  4980,  divide 

the  remainder  by  8,  and  subtract  123  from  the  quotient,  we  shall  find 
a remainder  equal  to  the  number  itself.  Ans.  444. 

21.  A laborer  engaged  for  40  days  on  these  conditions;  that  for 

every  day  he  worked  he  should  receive  80  cents,  but  for  every  day  he 
was  idle  he  should  forfeit  32  cents.  At  the  end  of  the  time  he  was 
entitled  to  $15.20  ; how  many  days  did  he  work,  and  how  many  was 
he  idle  ? Ans.  25,  and  15  days. 

22.  A cistern  containing  820  gallons  is  filled  in  20  minutes  by  3 
pipes,  the  first  of  which  conveys  10  gallons  more,  and  the  second  5 
gallons  less  than  the  third,  per  minute.  How  much  flows  through 
each  pipe  in  a minute?  Ans.  22,  7,  and  12  gallons. 

23.  A trader  maintained  himself  for  3 years  at  an  expense  of  £50 
a year,  and  each  year  augmented  that  part  of  his  stock  which  was  not 
thus  expended  by  g-  thereof.  At  the  end  of  the  third  year  his  original 
stock  was  doubled  ; what  was  that  stock  ? Ans.  £740. 


88 


SIMPLE  EQUATIONS. 


24.  A and  B began  to  trade  with  equal  sums  of  money.  The  first 

year  A gained  $40,  and  B lost  $40  ; the  second  year  A lost  i of  what 
he  had  at  the  end  of  the  first,  and  B gained  $40  less  than  twice  what 
A lost ; when  it  appeared  that  B had  twice  as  much  money  as  A. 
What  sum  did  each  begin  with?  Ans.  $320. 

25.  What  fraction  is  that,  whose  numerator  being  doubled,  and 

denominator  increased  by  7,  the  value  becomes  ; but  the  denomina- 
tor being  doubled,  and  the  numerator  increased  by  2,  the  value  be- 
comes f ? AllS.  |. 

26.  A and  B together  can  perform  a piece  of  work  in  8 days,  A 
and  C in  9 days,  and  B and  C in  10-days.  How  many  days  would  it 
take  each  person  to  perform  the  same  work  alone  ? 


Let  x,  y,  and  z represent  the  number  of  days  required  for  A,  B,  and 
C respectively ; 

Then  — is  the  part  of  the  work  that  A could  do  in  1 day,  &c. , 
and,  by  the  conditions  of  the  problem,  the  equations  will  be 


1 

x 


l 

8 > 


— + — = — 
y + z 10' 


By  subtracting  the  second  equation  from  the  first,  we  shall  elirm-  • 
nate  x,  and  then  by  adding  the  third  equation  we  shall  eliminate  z 

Ans.  A 14f-|  days,  B 17^,  C 23^-x. 

27.  From  two  places,  which  are  154  miles  apart,  two  persons  set 

out  at  the  same  time  to  meet  each  other,  one  traveling  at  the  rate  of  , 
3 miles  in  2 hours,  and  the  other  at  the  rate  of  5 miles  in  4 hours  ; in 
how  many  hours  will  they  meet  ? Ails.  56  hours. 

28.  In  a naval  engagement,  the  number  of  ships  captured  was  7 

more,  and  the  number  burned  was  2 less,  than  the  number  sunk.  Fif-  a 
teen  escaped,  and  the  fleet  consisted  of  8 times  the  number  sunk ; of 
how  many  ships  did  the  fleet  consist  ? Ans.  32. 

29.  A and  B together  could  have  completed  a piece  of  work  in  15 

days,  but  after  laboring  together  6 days,  A was  left  to  finish  it  alone,  i 
which  he  did  in  30  days.  In  how  many  days  could  each  have  per- 
formed the  work  alone?  Ans.  50,  and  21y  days. 

30.  On  comparing  two  sums  of  money  it  is  found,  that  ? of  the 
first  is  $96  less  than  of  the  second,  and  that  ■§■  of  the  secoud  is  as 
much  as  of  the  first.  What  are  the  sums  ? 

$720,  and  $512. 

31.  A privateer,  running  at  the  rate  of  10  miles  an  hour,  discovers 

a ship  18  miles  off,  making  way  at  the  rate  of  8 miles  an  hour.  In 
how  many  hours  will  the  ship  be  overtaken  ? Ans.  9 hours. 


SIMPLE  EQUATIONS. 


89 


32.  In  a composition  of  copper,  tin,  and  lead,  1 of  the  whole  minus 
16  pounds  was  copper,  ^ of  the  whole  minus  12  pounds  was  tin,  and  ~ 
of  the  whole  plus  4 pounds  was  lead  ; what  quantity  of  each  was  there 
in  the  composition?  Ans.  Copper  128,  tin  84,  lead  76  pounds. 

33.  The  sum  of  $660  was  raised  for  a certain  purpose  by  four  per- 
sons, the  first  giving  ^ as  much  as  the  second,  the  third  as  much  as 
the  first  and  second,  and  the  fourth  as  much  as  the  second  and  third. 
What  were  the  several  sums  contributed  ? 

Ans.  $60,  $120,  $180,  $300. 

34.  Two  pedestrians  start  from  the  same  point,  and  go  in  the  same 
direction  ; the  first  steps  twice  as  far  as  the  second,  but  the  second 
makes  3 steps  while  the  first  is  making  2.  How  far  has  each  one 
gone  when  the  first  is  300  feet  in  advance  of  the  second  ? 

Ans.  1200,  and  900  feet. 

35.  A merchant  has  cloth  at  $3  a yard,  and  another  kind  at  $5  a 

yard.  How  many  yards  of  each  kind  must  he  sell,  to  make  100  yards 
which  shall  bring  him  $450  ? Ans.  25,  and  75  yards. 

36.  In  the  composition  of  a quantity  of  gunpowder,  the  nitre  was 

10  pounds  more  .than  ■§■  of  the  whole,  the  sulphur  4-J  pounds  less  than 
•g-  of  the  whole,  and  the  charcoal  2 pounds  less  than  J-  of  the  nitre. 
What  was  the  amount  of  gunpowder?  Ans.  69  pounds. 

37.  Four  places  are  situated  in  the  order  of  the  letters  A,  B,  C,  D. 
The  distance  from  A to  D is  34  miles ; the  distance  from  A to  B is  § 
of  the  distance  from  C to  D ; and  ^ of  the  distance  from  A to  B,  plus 

of  the  distance  from  C to  D,  is  3 times  the  distance  from  B to  C. 
What  are  the  distances  between  A and  B,  B and  C,  C and  D ? 

Ans.  12,  4,  and  18  miles. 

38.  A vintner  sold  at  one  time  20  dozen  of  port  wine,  and  30  of 

sherry,  for  $120  ; and  at  another  time  30  dozen  of  port,  and  25  of 
sherry’  at  the  same  prices  as  before,  for  $140.  What  was  the  price  of 
a dozen  of  each  sort  of  wine  ? Ans.  $3,  and  $2. 

39.  A person  pays,  at  one  time,  to  two  creditors,  $53,  giving  to 

one  of  them  jj  of  the  sum  due  to  him,  and  to  the  other  $3  more  than 
| of  his  debt  to  him.  At  another  time  he  pays  them  $42,  giving  to  the 
first  of  what  remains  due  to  him,  and  to  the  other  ^ of  what  remains 
due  to  him.  What  were  the  debts  ? Ans.  $121,  and  $36. 

40.  A farmer  has  86  bushels' of  wheat  at  4s.  6 d.  per  bushel,  with 
which  he  wishes  to  mix  rye  at  3s.  6 d.  per  bushel,  and  barley  at  3s. 
per  bushel,  so  as  to  make  136  bushels,  that  shall  be  worth  4s.  a bushel. 
What  quantity  of  rye  and  of  barley  must  he  take  ? 

Ans.  14,  and  36  bushels. 


5* 


90 


SIMPLE  EQUATIONS 


41.  A composition  of  copper  and  tin,  containing  100  cubic  inches, 

weighs  505  ounces.  How  many  ounces  of  each  metal  does  it  contain, 
supposing  the  weight  of  a cubic  inch  of  copper  to  be  5j-  ounces,  and  of 
a cubic  inch  of  tin  4j  ounces  ? Ans.  420,  and  85  ounces. 

42.  A General  having  lost  a battle,  found  that  he  had  only  one- 

half  of  his  army  plus  3600  men  left,  fit  for  action;  of  his'  men  plus 
600  being  wounded,  and  the  rest,  who  were  J of  the  whole  army, 
either  slain,  taken  prisoners,  or  missing.  Of  how  many  men  did  his 
army  consist?  Ans.  24000. 

43.  Two  pipes,  one  of  them  running  5 hours,  and  the  other  4, 

filled  a cistern  containing  330  gallons  ; and  the  same  two  pipes,  the 
first  running  2 hours,  and  the  second  3,  filled  another  cistern  contain-  I 
ing  195  gallons.  How  many  gallons  did  each  pipe  discharge  pel 
hour  ? Ans.  30  and  45  gallons. 

44.  After  A and  B had  been  employed  on  a piece  of  work  for  14 

days,  they  called  in  C,  by  whose  aid  it  was  completed  in  28  days. 
Had  C worked  with  them  from  the  beginning,  the  work  would  have  | 
been  accomplished  in  21  days.  In  how  many  days  would  C alone 
have  accomplished  the  work  ? Ans.  42  days. 

45.  Some  smugglers  discovered  a cave  which  would  exactly  hold  J 

their  cargo,  viz.,  13  bales  of  cotton  and  33  casks  of  wine.  A revenue  I 
cutter  coming  in  sight  while  they  were  unloading,  they  sailed  away  i 
with  9 casks  and  5 bales,  leaving  the  cave  two-thirds  full.  Howmanv 
bales  or  casks  would  it  contain  ? Ans.  24  bales  or  72  casks. 

46.  A gentleman  left  a sum  of  money  to  be  divided  among  four 
servants,  so  that  the  share  of  the  first  was  1,  the  sum  of  the  shares  of 
the  other  three  ; the  share  of  the  second  -j  of  the  sum  of  the  other 
three  ; and  the  share  of  the  third  ^ of  the  sum  of  the  other  three  ; and 
it  was  also  found  that  the  share  of  the  first  exceeded  that  of  the  iast  by 
$14.  What  was  the  whole  sum  ? and  the  share  of  each  ? 

Ans.  Whole  sum  $120  ; shares  $40  ; $30;  $24,  $26. 


91 


CHAPTER  VI. 

RATI  0— P R 0 P 0 R T I 0 N— Y A R I A T I 0 N . 

RATIO. 

(127.)  The  ratio  of  one  quantity  called  the  antecedent  to  another 
of  the  same  kind  called  the  consequent , is  the  quotient  of  the  former 
divided  by  the  latter. 

Thus  the  ratio  of  12  to  4 is  3,  since  12  is  3 times  4 ; 
and  the  ratio  of  5 to  13  is  ^g-,  since  5 is  five  thirteenths  of  13. 

The  antecedent  and  consequent  together  are  called  the  terms  of  the 
ratio. 


Sign  of  Ratio. 

(128.)  A colon  ( : ) between  two  quantities  denotes  that  the  two 
quantities  are  taken  as  the  antecedent  and  consequent  of  a ratio. 

Thus  3 : 5,  the  ratio  of  3 to  5 ; a : b,  the  ratio  of  a to  b. 

(129.)  The  value  of  a ratio  may  always  be  represented  by  making 
the  antecedent  the  numerator , and  the  consequent  the  denomirvztor  ol 
a Fraction. 

Thus  3 : 5 is  equal  to  --  ; and  a : b is  equal  to  ~ , (75). 

Direct  and  Inverse  Ratio. 

(130.)  The  direct  ratio  of  the  first  of  two  quantities  to  the  second, 
is  the  quotient  of  the  first  divided  by  the  second  ; thus  the  direct  ratio 
of  3 to  5 is  y. 

The  inverse  ratio  of  the  first  quantity  to  the  second,  is  the  direct 
ratio  of  the  seco?id  to  the  first ; — in  other  words,  it  is  the  direct  ratio  of 
the  reciprocals  of  the  two  quantities. 

Thus  the  inverse  ratio  of  3 to  5 is  | ; — 
or  it  is  the  ratio  of  ^ to  §,  equal  to  equal  to  |-,  (127). 

Hence  inverse  is  often  called  reciprocal  ratio.  The  term  ratio 
used  alone  always  means  direct  ratio. 

E* 


92 


RATIO 


Compound  Ratio. 

% 

(131.)  A compound  ratio  is  the  ratio  of  the  product  of  two  or  more 
antecedents  to  the  product  of  their  consequents  ; and  is  equal  to  the 
product  of  all  the  simple  ratios. 

The  compound  ratio  of  a and  b to  x and  y is  — ; 

xy 

= the  product  of  the  simple  ratios  — and  — , or  — and  — . 

x y y x 

(132.)  The  ratio  of  the  first  to  the  last  of  any  number  of  quantities, 
is  equal  to  the  product  of  the  ratios  of  the  first  to  the  second,  the  second 
to  the  third,  and  so  on  to  the  last ; that  is,  it  is  compounded  of  all  the 
interve?iing  ratios. 

For  example,  take  the  quantities  a,  b,  x,  y.  The  ratios  of  the  first 
to  the  second,  the  second  to  the  third,  &c.,  are 

a b x . . abx  a . . , . 

— , — , — ; and  their  product  is  — , = — , which  is  a : y. 

o x y bxy  y 

Duplicate  and  Triplicate  Ratios. 

(133.)  The  duplicate  ratio  of  two  quantities  is  the  ratio  of  their 
squares , and  the  triplicate  ratio  is  the  ratio  of  their  cubes. 

Thus  the  duplicate  ratio  of  a to  b is  the  ratio  of  a2  to  b 2 ; 
and  the  triplicate  ratio  of  a to  b is  the  ratio  of  a3  to  b3 . 

The  subduplicatc  ratio  of  quantities  is  the  ratio  of  their  square 
roots,  and  the  subtriplicate  ratio  is  the  ratio  of  their  cube  roots 

Equimultiples  and  Equisubmultiples. 

(134.)  Equimultiples  of  two  quantities  are  the  products  which 
arise  from  multiplying  the  quantities  by  the  same  integer , and  equi- 
submultiples are  the  quotients  which  arise  from  dividing  the  quantities 
by  the  same  integer. 

Thus  3 a and  3 b are  equimultiples  of  a and  b,  while,  conversely,  a 
and  b are  equisubmultiples  of  3a  and  36. 

(135.)  Equimultiples,  or  equisubmultiples , of  two  quantities  have 
the  same  ratio  as  the  quantities  themselves,  (SI). 


PROPORTION. 


93 


I 


PROPORTION. 


(136.)  Proportion  consists  in  an  equality  of  the  ratios  of  two  or 
more  antecedents  to  their  respective  consequents — hut  is  usually  con- 
fined to  four  terms. 

(137.)  Four  quantities  are  in  Proportion  when  the  ratio  of  the 
first  to  the  second  is  equal  to  the  ratio  of  the  third  to  the  fourth  ; that 
is,  when  the  first  divided  by  the  second  is  equal  to  the  third  divided 
by  the  fourth. 

Thus  the  numbers  6,  3,  8,  4 are  in  proportion, 
since  the  ratio  equals  the  ratio 

And  the  quantities  a,  b,  x,  y are  in  proportion, 

CL  X 

when  the  ratio  — equals  the  ratio  — , (129). 


The  first  and  third  terms  are  the  antecedents  of  the  ratios ; the 
second  and  fourth  are  the  consequents.  The  first  and  fourth  are  the 
two  extremes  ; the  second  and  third  are  the  two  means. 

The  fourth  term  is  called  a fourth  'proportional  to  the  other  three 
taken  in  order ; thus  4 is  a fourth  proportional  to  6,  3,  and  8. 

(138.)  Three  quantities  are  in  Proportion  when  the  ratio  of  the 
first  to  the  second  is  equal  to  the  ratio  of  the  second  to  the  third , — the 
second  term  being  called  a mean  proportional  between  the  other  two. 

Thus  the  numbers  8,  4,  2 are  in  proportion, 
since  the  ratio  equals  the  ratio  ; 
and  4 is  a mean  proportional  between  8 and  2. 

Direct  and  Inverse  Proportion. 


(139.)  A direct  Proportion  consists  in  an  equality  between  two 
direct  ratios,  and  an  inverse  or  reciprocal  Proportion  in  an  equality 
between  a direct  and  an  inverse  ratio. 

Thus  the  numbers  6,  3,  8,  4 are  in  direct  proportion  ; (137). 

The  same  numbers  in  the  order  6,  3,  4,  8 are  in  inverse  proportion, 
since  the  direct  ratio  f is  equal  to  the  inverse  ratio  -§-,  (130). 


The  term  proportion  used  alone  always  means  direct  proportion. 


94 


PROPORTION— VARIATION. 


Sign  of  Proportion. 

(140.)  A Proportion  is  denoted  by  a double  colon  (:  :),  or  the  sign 
= between  the  equal  ratios  of  the  proportion. 

Thus  6 : 3 : : 8 • 4,  or  G : 3 = 8 : 4,  or  =| 
denotes  that  these  numbers  are  in  proportion,  and  is  read 
6 is  to  3 as  8 is  to  4. 

To  denote  an  inverse  Proportion  we  employ  the  sign  between 
the  two  ratios  of  such  proportion. 

Thus  6 : 3yt4  : 8,  denotes  that  6 is  to  3 inversely  as  4 is  to  8. 

Inverse  Converted  Into  Direct  Proportion. 

(141.)  An  inverse  is  converted  into  a direct  Proportion  by  inter- 
changing either  antecedent  and  its  consequent ; or  by  substituting  the 
reciprocals  of  either  antecedent  and  its  consequent. 

Thus  from  the  inverse  proportion  6 : 3 + 4 : 8, 
we  get  the  direct  proportion  3 : 6 = 4 : 8,  by  interchanging  6 and  3, 

or  1 : ^-=4  : 8,  by  substituting  A and 

The  reason  of  this  is  evident  from  the  nature  of  inverse  ratio,  (130). 

Variation. 

(142  ) Variation  is  such  a dependence  of  one  term  or  quantity  on 
another,  that  any  new  value  of  one  of  them  will  produce  a new  value 
of  the  other,  in  a constant  ratio  of  increase  or  diminution. 

1.  One  quantity  varies  directly  as  another  when  their  dependence  • 
is  such  that  if  one  of  them  be  multiplied,  the  other  must  be  multiplied, 
by  the  same  quantity. 

For  example,  the  Interest  on  money,  for  a given  time  and  rate  per  | 

cent.,  varies  dir  eel ly  as  the  Principal,  since  the  Interest  will  be  dou-  j 

bled,  or  tripled,  Ac.,  if  the  Principal  be  doubled,  or  tripled,  Ac. 

2.  One  quantity  varies  inversely  as  another  when  their  dependence 
is  such  that  if  one  of  them  be  multiplied,  the  other  must  be  divided, 
by  the  same  quantity. 

For  example,  the  Time  in  which  a given  amount  of  interest  will 
accrue  on  a given  principal,  varies  inversely  as  the  Rate  pet  cent., 
since  the  Time  will  be  doubled,  Ac.  if  the  Rate  be  halved,  Ac. 


VARIATION. 


95 

(143.)  When  one  quantity  varies  inversely  as  another,  the  product 
of  the  two  is  always  the  same  constant  quantity. 

For  as  one  of  the  two  quantities  is  multiplied , the  other  is  divided 
by  the  same  number ; the  product  of  the  two  will  therefore  be  multi- 
plied and  divided  by  the  same  number  ; hence  its  value  will  remain 
unchanged. 


Variation — an  Abbreviated  Proportion. 

(144.)  The  two  terms  of  a variation  are  the  two  antecedents  in  a 
Proportion  in  which  the  two  consequents  are  not  expressed,  but  may 
be  understood,  to  complete  the  proportion. 

Thus  when  we  say  that  the  Interest  varies  as  the  Principal,  for  a 
given  time  and  rate  per  cent.,  it  is  understood,  that 

The  Interest  on  any  principal  is  to  the  Interest  on  any  other  Prin- 
cipal, for  the  same  time  and  rate,  as  the  first  Principal  is  to  the 
second. 

Instead  of.  saying  “the  Interest  varies  as  the  Principal,”  we  may 
say,  the  Interest  is  proportional  to  the  Principal ; which  is  a brief 
method  of  expressing  a Proportion  by  means  of  its  antecedents , — the 
consequents  being  understood. 

- Sign  of  Variation. 

The  character  ~ placed  between  two  terms,  denotes  that  one 
of  them  varies  as  the  other.  Thus  x^y,  x varies  as  y,  or  x is 
proportional  to  y. 

a-~  — denotes  that  x varies  as  the  reciprocal  of  y \ or  x varies  re- 

s' 

ei [.locally  or  inversely  as  y. 

V 

— denotes  that  x varies  directly  as  y,  and  inversely  as  z \ 
z 

1 that  is,  x varies  as  the  quotient  of  y divided  by  z. 


96 


THEOREMS  IN  PROPORTION. 


Theorems  in  Proportion. 

(145.)  A Theorem  is  a proposition  to  be  demonstrated  or  proved. — 

A Corollary  is  an  inference  drawn  from  a preceding  proposition  01 
demonstration. 


Theorem  I. 


(146.)  Two  Fractions  having  a common  denominator , are  to  eacl 
other  as  their  numerators  ; and  two  fractions  having  a common  nume- 
rator are  to  each  other  inversely  as  their  denominators 
First.  Let  d he  the  common  denominator ; 

a c . 

then  the  ratio  of  — to  — is 
d d 


a c ad 

d ' d cd 


a 

c 


> (127), 


and  — - is  the  ratio  of  the  numerator  a to  the  numerator  c. 
c 

Secondly.  Let  n be  the  common  numerator ; 

, , . . n n . 

then  the  ratio  oi  — to  — is 
a c 

n n cn  c 

a c an  a ’ 

Q 

and  — is  the  inverse  ratio  of  the  denominator  a to  the  denomirtatoi 
a 

c,  (130). 

Therefore,  two  fractions  having  a common  denominator,  Ac. 

(147.)  Corollary.  The  value  of  a Fraction  varies  directly  as  its 
numerator , and  inversely  as  its  denominator. 


Theorem  II. 

(148.)  In  any  Proportion,  if  one  antecedent  he  greater  than  its  con- 
sequent, the  other  antecedent  will  be  greater  than  its  consequent ; il 
equal,  equal ; and  if  less,  less. 

Let  a : b : : x : y ; 

then  — = — , (lo/). 

by 

-Now  if  a be  greater  than  b the  first  ratio  will  be  greater  than  a 
unit,  and  consequently  the  second  ratio  will  he  greater  than  a unit, 
and  therefore  x will  be  greater  than  y.  In  like  manner  if  a he  equal 
to  b,  x will  be  equal  to  y , Ac. 

Hence,  in  any  Proportion,  if  one  antecedent,  Ac. 


THEOREMS  IN  PROPORTION. 


97 


Theorem  III. 

(149.)  When  four  quantities  are  in  Proportion,  the  product  of  the 
two  extremes  is  equal  to  the  'product  of  the  two  means. 

Let  a : b : : x : y ; 
then  is  ay=bx. 

For  since  the  quantities  are  in  proportion, 

= (137). 

by 

Clearing  the  Equation  of  fractions. 
ay  — bx . (115). 

Therefore,  when  four  quantities  are  in  Proportion,  &c. 

(150.)  Cor.  1.  A fourth  proportional  to  three  given  quantities,  is 
f uid  by  dividing  the  product  of  the  second  and  third  by  the  first. 

bx 

Thus  from  the  equation  ay  — bx , we  find  y=  — . 

(151.)  Cor.  2.  When  three  quantities  are  in  Proportion,  the  pro- 
duv  t of  the  two  extremes  is  equal  to  the  square  of  the  mean. 

For  let  a : b : : b : x ; then  ax=bb  = b2 . 

(152.)  Cor.  3.  A mean  proportional  between  two  given  quantities, 
is  tqual  to  the  square  root  of  their  product. 

Thus  from  the  equation  ax=b2,  we  find  b — [ax)2 . 

Theorem  IY. 


(153.)  When  the  product  of  two  quantities  is  equal  to  the  product 
of  two  other  quantities,  either  pair  of  factors  ?nay  be  made  the  extremes , 
and  the  other  the  means , of  a Proportion. 

Let  ab=xy ; 
then  will  a : x : : y : b. 

Dividing  both  sides  of  the  given  equation  by  b, 
xy 

a=T‘ 

Dividing  both  sides  of  this  last  equation  by  x, 

_ y_ 
x b 

Hence  a : x : : y : b.  (137). 

In  like  manner  a and  b may  be  taken  for  the  means , and  x and  y 
for  the  extremes.  Therefore,  when  the  product  of  two  quantities,  &e 


98 


THEOREMS  IN  PROPORTION. 


Theorem  V. 

(154.)  If  three  quantities  are  in  Proportion,  the  first  will  he  to  the 
third  as  the  square  of  the  first  to  that  of  the  second,  or  the  square  of 
the  second  to  that  of  the  third. 

Let  a : b : : b : x ; 

then  will  a : x ::  a2  : b2  ; or  a : x : : b2  : x2 . 

From  the  given  proportion,  we  find 
ax—b 2,  (151). 

Multiplying  both  sides  of  this  equation  by  a, 
a2x—ab2. 

Converting  this  equation  into  a Proportion, 
a : x : : a2  : b2 , (153). 

And  by  multiplying  both  sides  of  the  first  equation  by  x , we 
shall,  in  like  manner,  find  a : x : : b2  : x2. 

Therefore,  if  three  quantities  are  in  proportion,  &c. 

Theorem  YI. 

(155.)  Four  quantities  in  Proportion  are  also  in  proportion  by  in- 
version,— that  is,  when  each  antecedent  and  its  consequent  are  inter- 
changed with  each  other. 

Let  a : b : : x : y ; 
then  is  b : a : : y : x. 

From  the  given  proportion  we  find 
ay=bx,  (149). 

Making  h and  x the  extremes,  and  a and  y the  means, 
b : a ::  y : x,  (153). 

Hence,  four  quantities  in  proportion  are  also  in  proportion,  See. 
Theorem  VII. 

(15G.)  Four  quantities  in  Proportion  are  also  in  proportion  by  alter- 
nation,— that  is,  when  the  two  means,  or  the  two  extremes,  are  inter- 
changed with  each  other. 

Let  a : b ::  x : y; 

then  is  a : x ::  b : y;  or  y : b : : x : a. 

From  the  given  proportion  we  find 
ay—bx,  (149). 

This  equation  may  be  converted  into  the  proportion 
a : x : : b : y,  or  y : b ::  x : a,  (153). 

Therefore,  four  quantities  in  proportion,  Sec. 


THEOREMS  ET  PROPORTION. 


99 


Theorem  YIIT. 

(]  57.)  In  any  Proportion,  if  the  tivo  antecedents , or  the  two  conse- 
sequents,  or  an  antecedent  and  its  consequent,  be  multiplied  by  the 
same  quantity , the  products  and  the  remaining  terms  will  he  in  pro- 
portion. • 

Let  a \ b : : x : y ; 

and  let  n he  any  numerical  quantity  ; 
then  will  an  : b : : nx  : y,  &c. 

From  the  given  proportion  we  have 
ay—bx. 

Multiplying  hoth  sides  of  this  equation  hy  n, 
any—bnx. 

Converting  this  equation  into  a Proportion, 
an  : b : : nx  : y ; or  a : bn  : : x : ny  ; or  a : b : : nx  : ny,  (153). 

Therefore,  in  any  proportion,  if  the  two  antecedents,  &c. 

(158.)  Cor.  If  the  two  antecedents , or  the  two  consequents,  or  an 
antecedent  and  its  consequent,  be  divided  by  the  same  quantity,  the 
quotients  and  the  remaining  terms  will  be  in  proportion. 

For  dividing  by  a quantity  is  equivalent  to  multiplying  by  its  re- 
ciprocal. 

Theorem  IX. 

(159.)  Four  quantities  in  Proportion  are  also  in  proportion  by  com- 
position,— that  is,  the  sum  of  the  first  and  second  terms  is  to  the  first  or 
second,  as  the  sum  of  the  third  and  fourth  is  to  the  third  or  fourth 
Let  a : b ::  x : y 
then  is  a-\-b  : a ::  x-fiy  : x. 

From  the  given  proportion  we  have 
ay=bx. 

Adding  both  sides  of  this  equation  to  ax, 
ax + ay = ax + bx . 

Resolving  each  member  of  this  equation  into  its  factors, 
a{x+y)=.x{a+b). 

Converting  this  last  equation  into  a Proportion, 
a-\-b  : a : : x-\-y  : x,  (153). 

By  adding  both  sides  of  the  first  equation  to  by,  it  may  be  proved, 
n like  manner,  that  a-\-b  : b : : x-\-y  : y. 

Therefore,  four  quantities  in  proportion  are  also  in  proportion,  &c 


100 


THEOREMS  IN  PROPORTION. 


Theorem  X. 

(160.)  Four  quantities  in  Proportion  are  also  in  proportion  by  dim 
sion, — that  is,  the  difference  of  the  first  and  second  terms  is  to  the  first 
or  second,  as  the  difference  of  the  third  and  fourth  is  to  the  third  or 
fourth. 

Let  a : b : : x : y ; 
then  is  a — b : a : : x — y : x. 

From  the  given  proportion  we  find 
ay=hx. 

Subtracting  both  sides  of  this  equation  from  ax, 
ax — ay=ax — bx. 

Resolving  each  member  of  this  equation  into  its  factors, 
a(x—y)=x(a—b). 

Converting  this  last  equation  into  a Proportion, 
a— b : a : : x—y  : x. 

By  subtracting  both  sides  of  the  first  equation  from  by,  it  may  b* 
proved,  in  like  manner,  that  a — b : b : : x — y : y. 

Hence,  four  quantities  in  proportion  are  also  in  proportion,  &c. 


Theorem  XI. 

(161.)  When  any  number  of  quantities  are  in  Proportion,  the  sum 
of  any  two  or  more  of  the  antecedents  is  to  the  sum  of  their  conse- 
quents, as  any  one  antecedent  is  to  its  consequent. 

Let  a : b : : c : d : : x : y,  &cc.  ; 
then  is  a-\-c  : b-\-d  : : x : y. 

From  the  given  proportion  we  shall  find 
ay—bx,  and  cy—dx,  (149). 

Adding  together  the  corresponding  members  of  these  two  equations, 
aiy-\-cy=.bx-\-dx. 

Resolving  each  member  of  this  equation  into  its  factors, 

( a + c)y=(b+d)x , 

Converting  this  equation  into  a Proportion, 
a-\-c  : bfd  : : x : y. 

By  adding  xy  to  both  sides  of  the  third  equation,  we  shall,  in  like 
manner,  find  that  a + c+a?  : b-\-d+y  : : x : y. 

Therefore,  when  any  number  of  quantities  are  in  proportion,  &c. 


THEOREMS  IN  PROPORTION. 


10] 


Theorem  XII. 

(162. ) If  two  Proportions  have  an  antecedent  and  its  consequent , or 
the  two  antecedents,  or  the  two  consequents,  the  same  in  both , the  re- 
maining terms  will  be  in  proportion. 

Let  a : b : : x : y, 
and  a : b : :w: z ; 
then  will  x : y : : w : z. 

For  the  ratio  of  x to  y is  equal  to  the  ratio  of  w to  z,  since  each  of 
these  ratios  is  equal  to  the  ratio  of  a to  b ; hence 

x \y  : : w:  z. 

The  two  given  proportions  have  an  antecedent  and  its  consequent 
the  same  in  both.  If  the  two  antecedents  were  the  same  in  both,  the 
demonstration  would  be  the  same,  after  interchanging  the  means; 
and  if  the  two  consequents  were  the  same,  after  interchanging  the  ex- 
tremes, (156). 

Hence,  if  two  proportions  have  an  antecedent  and  its  consequent, 
&c. 

Two  or  more  Proportions  having  an  antecedent  and  its  consequent 
the  same  in  each,  form  one  continued  proportion. 

Thus,  a : b ::  u : v, 
a \b  ::  iv : x, 
and  a : b ::  y : z, 

form  the  continued  proportion,  a : b ::  u : v ::  w : x : y : z. 

Theorem  XIII. 

(163.)  The  sum  of  the  first  and  second  terms  in  any  Proportion,  is 
to  their  difference,  as  the  sum  of  the  third  and  fourth  is  to  their  differ- 
ence. 

Let  a : b : : x : y, 

then  is  a-\-b  : a — b : : x-\-y  : x — y. 

By  Composition  and  Division,  in  the  given  proportion. 

a-\-b  : a : : x-\-y  : x;  (159)  ; 
a — b : a : : x—y  : x.  (160). 

These  two  proportions  have  the  consequents  a and  x the  same  ir, 
both;  hence  a-{-b  : a—b  : : x-\-y  : x — y,  (162). 

Therefore,  the  sum  of  the  first  and  second  terms  in  any  proportion, 
&c. 


102 


THEOREMS  IN  PROPORTION. 


Theorem  XIY. 

(164.)  The  •products  of  the  corresponding  terms  of  two  or  more 
Proportions,  are  in  proportion. 

Let  a : b ::  x : y, 
and  c : d : : w : z ; 
then  is  ac  : bd  : : xw  : yz. 

From  the  two  given  proportions,  we  have 
ay  — bx,  and  cz=dw. 

Multiplying  together  the  corresponding  members  of  these  equations. 
ac  .yz-Sbd . xw. 

Converting  this  equation  into  a Proportion, 
ac  : bd  : : xw  : yz,  (153). 

In  like  manner  the  demonstration  may  be  extended  to  three  or 
more  proportions.  Hence,  the  products  of  the  corresponding  terms,  &c. 

(165.)  Cor.  Like  poivers  or  roots  of  proportional  quantities,  are  in 
proportion. 

For  if  a : b : : x : y,  by  multiplying  each  term  by  itself,  we  shall 
have,  according  to  the  Theorem,  a2  : b2  : : x2  : y2  ; and  multiplying 
these  by  the  given  terms,  we  shall  have  a3  : b3  : : x3  : y3,  k c. 

Theorem  XY. 

(166.)  For  any  factors  in  an  antecedent  and  its  consequent,  or  the 
two  antecedents,  or  the  two  consequents,  in  a Proportion,  may  be  sub- 
stituted any  other  quantities  which  have  the  same  ratio  to  each  other 
Let  a : b : : nx  : py, 
and  n : p : : r : s ; 
then  will  a : b : : rx  : sy. 

From  the  two  given  proportions,  we  find 
an  \ bp  w nrx  : psy,  (164). 

Dividing  the  antecedents  in  this  proportion  by  n,  and  the  consequents 
by  P< 

a : b : : rx  : sy,  (1 58). 

This  proves  the  first  affirmation  in  the  Theorem.  By  inter- 
changing the  extremes  in  the  two  given  proportions,  and  afterwards 
the  means,  (156),  the  other  two  affirmations  in  the  theorem  may  be 
demonstrated. 

Therefore,  for  any  factors  in  an  antecedent  and  its  consequent,  kc 


! GENERAL  SOLUTIONS  OF  PROBLEMS  103 

(general  Solutions  of  Problems. — Formulas.  Applications 
of  Proportion,  &c. 

(167.)  In  the  general  solution  of  a Problem,  all  the  quantities  are 
represented  by  letters  ; and  the  unknown  being  thus  found  in  terms  of 
all  the  known  quantities,  the  result  discloses  a rule  for  the  numerical 
computation  in  any  given  case  of  the  problem. 

example. 

To  find  two  numbers  whose  sum  shall  be  s,  and  difference  d. 

Let  x represent  the  greater , and  y the  less  number. 

From  the  conditions  of  the  problem,  we  have 
x+y=s, 
and  x — y—d. 

Py  adding  the  second  equation  to  the  first ; and  also  subtracting 
the  second  from  the  first,  we  have 

2 x=  s+d,  and  2y=  s — d, 

, . , . s+d  s—d 

which  give  X— — and  y=  — ^ — . 

From  these  general  values  of  x and  y,  we  learn  that  the  greater  of 
two  numbers  is  equal  to  of  (the  sum  + the  difference ),  a,nd  that  the 
less  is  equal  to  of  (the  sum  — the  difference),  of  the  two  numbers ; 
hence  the  following  rule  : 


B(1G8.)  To  find  two  numbers  from  their  sum  and  difference, — 
Add  the  difference  to  the  sum,  and  divide  by  2,  for  the  greater  of  the 
two  numbers ; subtract  the  difference  from  the  sum,  and  divide  by  2, 
for  the  less  number. 

For  example,  if  the  sum  of  two  numbers  be  500,  and  their  differ- 
snce  146, 

m , . 500  + 146 

The  greater  number  is — — =323  ; 

2 

, . , , . 500-146 

and  the  less  number  is — ■ — =177. 

2 

1(169.)  An  algebraic  Formula  is  an  equation  between  the  symbols 
of  certain  quantities — resulting  from  the  general  solution  of  a Problem, 
or  the  investigation  of  some  general  principle. 

Thus  the  equations  which  express  the  values  of  x and  y in  the 
preceding  Example,  are  formulas  for  finding  two  numbers  from  the 
sum  and  difference  of  the  numbers. 


104 


PROBLEMS. 


PROBLEMS 

In  Proportion , Percentage , Interest,  Spc 

(170.)  A Proportion  occurring  in  the  solution  of  a Problcrs , may  bt 
converted  into  an  Equation  by  putting  the  product  of  the  two  extrema 
equal  to  the  product  of  the  two  means,  (149). 

EXAMPLE. 

To  divide  $1000  between  three  persons,  in  the  proportions  of  2,  3, 
and  5 ; that  is,  so  that  A’s  share  shall  be  to  B’s  as  2 to  3,  and  B’s  to 
C’s  as  3 to  5. 

Let  x represent  A’s  share  ; y,  B’s  share  ; and  2,  C’s  share.  Then 
by  the  conditions  of  the  problem,  we  have 

a:+y  + 2=1000  ; 

x : y : : 2:3; 
and  y : z : : 3 : 5. 

By  converting  the  two  Proportions  into  Equations,  we  find 
ox  — 2y,  and  5y=3z. 

We  have  now  three  equations  from  which  to  find  the  values  of  the 
three  unknown  quantities,  (125). 

The  solution  of  the  Problem  may  also  be  effected  with  one  unknown 
quantity,  by  finding  fourth  proportionals  for  the  shares  of  B and  C. 

Let  x represent  A’s  share  ; 

then  2 : 3 : : x : ; and  2 : 5 : : x : ^ # (150). 

Hence  B’s  share  is  represented  by  — , and  C’s  by  ~ ; and  the 
equation  of  the  problem  is 

3a:  5a:  _ 

x-\ — - — | — — =$1000. 

Ans.  The  several  shares  are  $200,  $300,  $500 

The  general  Problem  of  which  the  preceding  is  a particular  case, 
may  be  stated  thus ; — To  divide  the  sum  s between  three  persons  in 
the  proportions  of  a,  b,  and  c. 

The  Formulas  which  would  be  found  for  the  several  shares,  are 
as  bs  cs 

a-\-b-\-c’  a-f-5+c’  a-\-b-\-c' 

These  Formulas  translated  into  arithmetical  language,  would  fur- 
nish a Proposition,  or  a Rule,  which  might  be  applied  to  any  given 
case  of  the  general  problem. 

• 


PROBLEMS. 


105 


tions  of  ¥ 


It  may  bo  here  remarked  that  algebraic  Formulas  express  general 
principles  and  methods  of  solution  with  the  utmost  distinctness  and 
brevity, — but  that  it  is  not  always  possible  to  translate  them  into  con- 
cise and  perspicuous  phraseology. 

EXERCISES. 

1.  Divide  $950  between  two  persons  so  that  their  shares  shall  bo 
to  each  other  as  3 to  5. 

This  problem  might  be  solved  without  employing  proportion,  by  ob- 
serving that  the  first  share  will  be  § of  the  second. 

Ans.  $3561;  $593|. 

2.  Find  the  Formulas  for  dividing  any  given  sum  s between  two 
persons  so  that  the  shares  shall  be  to  each  other  as  any  two  numbers 

a and  b.  as  bs 

Ans.  and . 

«+»  a+b 

3.  Divide  the  sum  of  $3000  between  A,  B,  and  C,  in  the  propoi- 

tions  of  1,  2,  and  3.  Ans.  $500  ; $1000  ; $1500. 

4.  Divide  the  sum  of  $7600  between  three  persons,  in  the  propor- 

i and  l . Ans.  $4000;  $2000;  $1600. 

5.  A bankrupt  is  indebted  to  A $400,  and  to  B $700.  He  is  able 
to  pay  to  both  $900  ; what  sum  should  each  of  the  two  creditors  re- 
ceive ? 

The  $900  should  be  divided  in  the  proportion  of  400  and  700,  or, 
of  4 and  7,  (158).  Ans.  $327T3T  ; $572T8T. 

6.  Three  persons  engaged  in  a speculation  towards  which  they 
contributed,  respectively,  $300,  $400,  and  $500.  The  profit  amounted 
to  $550  ; what  are  the  respective  shares  of  profit  1 

Ans.  $137£;  $1831;  $2291. 

7.  A,  B,  and  C in  a joint  mercantile  adventure  lost  $742.  A’s 
part  of  the  capital  employed  was  to  B’s  as  4 to  3,  and  B’s  was  to  C’s 
as  5 to  6 ; what  amount  of  loss  should  be  borne  by  each  ? 

Ans.  $280;  $210;  $252. 

8.  Four  persons  rented  a pasture,  in  which  the  first  kept  8 oxen, 
the  second  6,  the  third  10,  and  the  fourth  12.  The  sum  paid  was  $40  ; 
what  amount  should  have  been  paid  by  each  person  1 

Ans.  $8|;  $6f ; fill;  $13$. 

9.  A testator  bequeathed  his  estate,  amounting  to  $7830,  to  his 
three  children,  in  such  a manner  that  the  share  of  the  first  was  to  that 
of  the  second  as  2\  to  2,  and  the  share  of  the  second  to  that-  of  the 

What  were  the  shares  1 

Ans.  $3150;  $2520;  $2160. 

10.  A,  B,  C,  and  D together  have  $3000  ; A’s  part  is  to  B’s  as  9 
to  3,  B and  C together  have  $1500,  and  C’s  part  is  to  D’s  as  3 to  4 
What  is  the  sum  possessed  by  each  person  ? 

Ans.  $500;  $750;  $750;  $1000 


third  as  3^  to  3. 


106 


PROBLEMS. 


11.  Three  persons  contributed  funds  in  a joint  speculation  as  foi 
lows  : A $200  for  5 months,  B $400  for  3 months,  and  C $500-  for  4 
months.  The  profit  amounted  to  $600  ; what  are  the  several  shares 
of  profit  ? 

Each  Dollar  contributed  produced  a Profit  •proportional  to  the 
Time  it  was  in  the  business.  Each  person’s  share  of  profit  is  therefore 
proportional  to  his  amount  of  capital  X its  time ; in  other  words,  the 
respective  shares  are  to  each  other  in  the  compound  ratio  of  capital 
and  time,  (131), 

Hence,  A s share  of  profit  is  to  B's  as  200  X 5 to  400  X 3 ; 
and  A’s  is  to  C’s  as  200  x 5 to  500  X 4. 

Dividing  the  antecedent  and  consequent  by  100,  these  ratios  become 
2 x 5 to  4 x3  and  2 x 5 to  5 x4,  (158). 

By  still  further  reductions,  on  the  same  principle,  the  ratios  become 
5 to  6,  and  1 to  2. 

Ans.  $142$;  $171$;  $285$. 

12.  Two  persons  rented  a pasture  for  $43.  The  first  put  into  it 
100  sheep  for  15  days,  and  the  second  120  sheep  for  9 days;  what 
amount  of  rent  should  be  paid  by  each  person  ? 

Ans.  $25  and  $18. 

13.  A,  B,  and  C trade  together;  A ventures  $1000  for  5 months, 

B $1200  for  4 months,  and  C $800  for  7 months.  The  profits  of  the 
partnership  amount  to  $2310  ; what  share  of  profit  should  be  assigned 
to  each  ? Ans.  $750;  S720  ; $840. 

14.  An  estate  consisting  of  1000  acres  of  land  is  to  be  divided  be- 
tween three  persons,  so  that  the  first  share  shall  be  to  the  second  as  2 
to  3,  and  the  first  to  the  third  as  1 to  2.  What  are  the  shares  ? 

Ans.  222-| ; 333$  ; 444$,  acres. 

15.  Two  men  contracted  to  do  a certain  work  for  $5000.  In  ac- 
complishing the  work,  the  first  employed  100  laborers  for  50  days  ; and 
the  second  125  laborers  for  60  days  ; — to  what  shares  of  the  stipulated 
sum  are  the  two  men  respectively  entitled  ? 

Ans.  $2000  ; and  $3000. 

16.  A gentleman  bequeathed  $18000  to  his  widow  and  his  three 
sons,  in  the  proportions  of  2,  2$,  3,  and  3$  respectively.  His  widow 
dying  before  the  division  was  effected,  the  whole  is  to  be  divided  pro 
portionably  among  the  three  sens.  What  are  their  several  shares  ? 

Ans.  $5000;  $6000;  $7000. 

17.  Find  the  Formulas  for  dividing,  between  two  partners,  the 
profits  s of  a joint  adventure,  in  which  the  first  had  the  Capital  a for 
the  Time  b,  and  the  second  the  Capital  c for  the  Time  d. 

. abs  , cds 

Ans.  — and  — 3 . 

ab-\-cd  ab-\-ca 


PROBLEMS. 


107 


Problems  in  Percentage. 

(171.)  Percentage  is  an  allowance  at  a certain  rate  per  hundred ; 
and  this  rate  is  called  the  rate  per  cent.,  from  the  Latin  centum , which 
means  a hundred. 

The  ratio  of  percentage  is  the  rate  per  cent,  h-  100,  and  is  there- 
fore equal  to  the  rate  per  unit. 

The  basis  of  percentage  is  the  sum  or  number  on  which  an  amount 
of  percentage  is  computed. 

From  these  definitions  it  follows,  that 

(172.)  The  basis  of  percentage  X the  ratio  of  percentage  produces 
the  amount,  of  percentage  ; and,  conversely,  that  the  amount  the 
ratio  produces  the  basis  of  percentage. 


HHP"  The  Student  may  be  required  to  write  the  Rules  of  Percent- 
age which  are  indicated  by  the  Formulas  among  the  following  prob- 
lems. 

18.  A merchant  finds  that  his  capital,  which  is  now  $12000,  has 
increased  in  one  year  at  the  rate  of  20  per  cent.  ; what  was  his  capital 
at  the  beginning  of  the  year? 

Let  x represent  his  capital  at  the  beginning  of  the  year  : 

then  100  : 120  : : a;  : 12000.  Ans.  $10000. 

19.  What  is  the  Formula  for  finding  a sum  of  money  which,  in- 
creased at  the  rate  of  r per  cent.,  shall  amount  to  the  sum  a. 


Ans. 


100a 


I-'"'  100 -|-r 

20.  An  agent  receives  $500  to  be  laid  out  in  merchandise,  after 
deducting  his  commission  of  1-^  per  cent,  on  the  amount  of  the  pur- 
jlbhase.  What  will  be  the  amount  of  the  purchase  ? 

Ans.  $492  61’. 

21.  A merchant  obtains  an  insurance  at  2 per  cent,  on  a stock  of 
pods  valued  at  $7500,  which  includes  this  amount  and  the  premium 
or  the  insurance.  What  is  the  sum  insured  ? Ans.  $7653.06’. 

22.  Wrhat  is  the  Formula  for  finding  a sum  of  money  which,  di- 
ninished  at  the  rate  of  r per  cent.,  shall  leave  the  sum  a? 

100  a 

Ans.  — — . 

1 00 — r 

23.  The  profits  of  a manufacturing  company  this  year  amount  to 
13096,  which  is  31  per  cent,  less  than  their  profits  last  year.  What 
jvas  the  amount  of  profits  last  year  ? Ans.  $3200. 

24.  What  must  be  the  percentum  of  profit  at  which  a quantity  of 


aerchandise,  bought  for  $3750,  must  be 
f profit  shall  be  $1500  ? 


?old,  that  the  whole  amount 
Ans.  40  per  cent. 


108 


PROBLEMS. 


25.  What  is  the  Formula  for  finding  the  rate  per  cent,  at  which 
the  sum  s must  be  increased  to  produce  the  sum  a 1 

100(a — s) 

Ans.  — — . 

s 

26.  A quantity  of  silk  was  purchased  for  $220,  and,  on  account  of 

its  having  become  damaged,  was  sold  for  $176.  What  was  the  per- 
centum  of  the  loss  sustained  1 Ans.  20  per  cent. 

27.  What  is  the  Formula  for  finding  the  rate  per  cent,  at  which 
the  sum  s must  be  diminished  to  leave  the  sum  a ? 

Am. 

s 

28.  Wrhat  amount  of  stock  in  an  Insurance  Office,  at  a discount  of 

5 per  cent.,  could  be  purchased  for  $3800  ? Ans.  $4000. 

29.  A merchant  finds  that  his  capital,  which  is  now  $4350,  has 

decreased  in  one  year  at  the  rate  of  12^  per  cent.  What  was  his  I 
capital  at  the  beginning  of  the  year  1 A)is.  $4971.42’. 

30.  What  amount  of  stock  in  a manufacturing  establishment,  at 
an  advance  of  6^  per  cent.,  could  be  purchased  for  $1200  ? 

Ans.  $1126.76’. 

31.  A quantity  of  damaged  cloth  wras  sold  for  $250. — which  was 
at  a loss  of  16|-  per  cent.  For  what  sum  was  the  cloth  purchased  1 

Ans.  $300.  I 

32.  The  population  of  a city  increased  from  7850  to  11775  inhabit- 

ants, in  one  year.  What  was  the  percentum  of  increase  during  the 
year  ? A?is.  50  per  cent. 

33.  An  agent  receives  $2030  to  invest  in  merchandise — himself  to  I 

retain  a commission  of  1^-  per  cent,  on  the  amount  of  the  purchase. 
What  is  the  sum  to  be  invested  1 Ans.  $2000.  | 

34.  A merchant  wishes  to  effect  an  insurance  on  a stock  of  goods, 

amounting  to  $3573,  which  shall  cover  both  the  value  of  the  goods 
and  the  premium  of  insurance.  What  is  the  sum  to  be  insured,  al- 
lowing the  rate  to  be  ■§•  per  cent  ? -d.«s.  $3600.  I 

35.  What  amount  of  stock  in  a Savings  Bank,  at  an  advance  of  c I 
per  cent.,  could  be  purchased  for  $4200  ? and  what  amount  in  another 
at  a discount  of  5 per  cent.,  could  be  purchased  for  $1995  ? 

$4000,  and  $2100.  j 


PROBLEMS. 


109 


Problems  in  Interest,  Spc. 

(173.)  Interest  is  the  price  or  premium  paid  for  the  use  of  money , 
and  is  reckoned  at  a certain  percentum,  annually , on  the  sum  for 
which  it  is  paid. 

The  Principal  is  the  sum  for  which  Interest  is  paid — the  Amount 
is  the  sum  of  the  Principal  and  Interest. 

From  the  principles  of  Percentage,  (172),  it  is  evident  that, 

' 

(174.)  The  Principal  X the  ratio  of  percentage  produces  the  inter- 
est, for  one  year.  For  the  ratio  of  percentage,  in  this  case,  is  the  inter- 
est of  $ 1 for  one  year. 


[CP“  The  Student  may  be  required  to  write  the  Rides  of  Interest 
which  are  indicated  by  the  Formulas  among  the  following  problems. 

36.  What  Principal  would  amount  to  $1000  in  5 years,  allowing 
the  rate  of  interest  to  be  6 per  cent  ? 

Let  x represent  the  Principal  required ; 


6 6x 
then  xX  -ttv  or 


is  the  Interest  for  one  year  ; 


100  100 
and  by  adding  5 years’  interest  to  the  Principal, 
Ox 


we  have  x-\- 


100 


— $1000. 


Ans.  $769^. 

37.  What  is  the  Formula  for  finding  the  Principal  which,  at  inter- 
est at  r per  cent.,  would  amount  to  the  sum  a in  t years? 

. 100a 

Ans-  TMTS- 

38.  What  Principal  would  amount  to  $2500,  in  10  years,  allow- 
ing the  rate  of  interest  to  be  7 per  cent.  ? Ans.  $1470.588’. 

39.  At  what  Fate  per  cent,  must  $1000  be  put  on  interest,  to 

amount  to  $1150  in  2 years  and  6 months?  Ans.  6 per  cent. 

• 40.  What  is  the  Formula  for  finding  the  Fate  per  cent,  of  interest 
at  which  the  sum  s would  amount  to  the  sum  a in  t years  ? 

100(a— s) 


Ans. 


st 


41.  In  how  many  years  would  $6000  amount  to  $7470,  allowing 
the  rate  of  interest  to  be  7 per  cent?  Ans.  3-|  years 


110 


PROBLEMS. 


42.  What  is  the  Formula  for  finding  the  Time  in  which  the  sum  s 
would  amount  to  the  sum  a,  if  the  interest  be  at  r per  cent  ? 

100<-s> . 

sr 


43.  What  Principal  would  produce  as  much  interest  in  3^  years, 
as  $500  would  in  4 years,  the  rate  of  interest  in  both  cases  being  6 
per  cent  1 Ans.  $57  ll . 


44.  At  what  Hate  per  cent,  of  interest  would  $525  pi'oduce  the 

same  amount  of  interest  in  5 years,  that  $700  would  produce  at  5 per 
cent,  in  3 years  1 Ans.  4 per  cent. 

45.  A person  who  possessed  a capital  of  $70000,  put  the  greater 

part  of  it  at  interest  at  5 per  cent.,  and  the  other  part  at  4 per  cent. 
The  interest  on  the  whole  was  $3250  per  annum  ; required  the  two 
parts.  Ans.  $45000,  and  $25000. 

46.  The  sum  of  $200  is  to  be  applied  in  part  towards  the  payment 
of  a debt  of  $300,  and  in  part  to  paying  the  Interest,  at  6 per  cent.,  in 
advance , for  12  months,  on  the  remainder  of  the  debt?  What  is  the 
amount  of  the  payment  that  can  be  made  on  the  debt  ? 

Let  x represent  the  payment ; 

then  (300 — *)  X j-g-Q  is  the  Interest  on  the  remainder  of  the  debt ; and 
we  have  therefore  the  Equation, 

a:  + (300 — x)  xT^o=200. 

Ans.  $193.61’. 

47.  A is  indebted  to  B $1000,  and  is  able  to  raise  but  $600.  With 

this  sum  A proposes  to  pay  a part  of  the  debt,  and  the  Interest,  at  3 
per  cent.,  in  advance,  on  his  Note  at  2 years  for  the  remainder.  For 
what  sum  should  the  note  be  drawn  ? Ans.  $476.19’. 


48.  Find  the  Formulas  for  dividing  the  sum  s into  two  parts,  one 
of  which  is  to  be  applied  towards  the  payment  of  a debt  of  n dollars, 
and  the  other  to  paying  the  interest,  in  advance,  on  the  remainder  of 
the  debt,  for  t years,  at  r per  cent,  per  annum. 


Ans. 


100  s—nrt  rt(n-s) 

and  — ™ 


100 — rt  ’ 100— rt 

What  would  be  the  Rule  for  finding  the  amount  of  payment  that 
could  be  made  on  the  debt  ? 


Ill 


CHAPTER  YII. 

ARITHMETICAL,  HARMONICAL,  AND  GEOMETRICAL  PROGRESSION. 


ARITHMETICAL  PROGRESSION. 


(175.)  An  arithmetical  progression  is  a series  of  quantities  which 
continually  increase  or  decrease  by  a common  difference. 

Thus  1.  3,  5,  7,  9,  is  a Progression  in  which  the  quantities  increase 
by  the  continual  addition  of  the  common  difference  2. 

And  15,  12,  9,  6,  3,  is  a progression  in  which  the  quantities  de- 
crease by  the  continual  suht-raction  of  the  common  difference  3. 

The  first  and  last  terms  of  the  Progression  are  called  the  two  ex- 
tremes, and  all  the  intermediate  terms  the  means. 

[ 

The  theory  of  Arithmetical  Progression  is  contained  in  the  follow- 
ing propositions. 

The  Last  Term. 

(176.)  The  last  term  of  an  increasing  Arithmetical  Progression,  is 
equal  to  the  first  term  + the  product  of  the  common  difference  X the 
number  of  terms  less  one;  and  in  a decreasing  Progression  it  is  equal 
to  the  first  term  — - the  same  product. 

Let  a he  the  first  term,  and  d the  common  difference  ; then  in  an 
increasing  progression  the  series  will  be, 

a , a-\-d,  a-\-2d,  a-\-c>d , a + 4c?,  &c.  ; 

and  in  a decreasing  progression  the  series  will  be, 

a,  a—d , a — 2d,  a — 3 d,  a — 4 d,  &c. 

I 

In  these  series  th e fifth  or  last  term  a ±^d,  a plus  or  minus  4 d,  is 
the  first  term  a plus  or  minus  4 times  the  common  difference  d.  And 
the  proposition  is  evidently  true  for  any  number  of  terms. 

(177.)  Cor.  The  common  difference  of  the  terms  in  an  Arithmetical 
Progression,  is  equal  to  the  difference  between  the  two  extremes  the 
number  of  terms  less  one. 


112 


PROGKESSION. 


The  Sum  of  the  Two  Extremes. 

(178.)  The  sum  of  the  two  extremes  in  an  Arithmetical  Progression, 
is  equal  to  the  sum  of  any  two  terms  equidistant  from  them , or  to 
twice  the  middle  term  when  the  number  of  terms  is  odd. 

Let  a be  the  first  term,  and  d the  common  difference  ; then  in  an 
increasing  progression  the  series  will  be, 

a , a-\-d,  a-\-2d,  «+3cZ,  a + 4r7,  &c. 

Of  these  five  terms  the  sum  of  the  first  and  the  last,  is 
a -}“  ( a -f  4 d')  = 2a 4 d. 

The  sum  of  the  second  and  the  fourth,  which  are  equidistant  from 
the  extremes,  is  ( a-\-d)-\-(a-\-3d ),  also  =2a-\-\d. 

We  see  moreover,  that  the  sum  of  the  two  extremes  is  equal  to 
twice  the  middle  term,  a + 2 d. 

In  like  manner  the  proposition  will  he  found  true  for  any  number 
of  terms  ; as  also  when  the  Progression  is  a decreasing  one. 

(179.)  Cor.  An  arithmetical  mean  between  two  given  terms,  is 
equal  to  half  the  sum  of  those  terms. 

For  the  sum  of  the  two  given  terms,  considered  as  the  two  extremes 
of  an  Arithmetical  Progression,  is  equal  to  twice  the  mean  or  middle 
term. 


The  Sum  of  all  the  Terms. 

(180.)  The  sum  of  all  the  terms  of  an  Arithmetical  Progression,  is 
equal  to  half  the  sum  of  the  two  extremes  X the  number  of  terms. 

To  prove  this  proposition  we  add  the  several  terms  of  an  Arithme- 
tical Progression  to  those  of  the  same  progression  reversed ; thus 

a,  a,-\-  d,  a-\~2d,  a-\-t$d, 

a-\-3d  a~\-2d  a-{~  d a 

2«+3c7’  2a-f-3c7’  2a-\-od’  2a-\-‘3d' 

The  sum  ( 2a.-\-od)-\-(2a-\-od ) &c.  of  the  tico  series,  is  the  sum  of 
the  two  extremes  in  either  series  X the  number  of  terms  ; hence  the 
sum  of  either  series  is  equal  to  half  the  sum  of  the  two  extremes  X 
the  number  of  terms.  The  demonstration  will  evidently  apply  to  any 
number  of  terms. 


PROGRESSION. 


113 


Formulas  in  Arithmetical  Progression. 

(181.)  In  an  Arithmetical  Progression,  let  a be  the  first  term,  d the 
common  difference,  n the  number  of  terms,  l the  last  term,  and  S the 
sum  of  all  the  terms.  Then 

A.  ...  . I =a±d(n—  1)  (176); 

B . . . . S=±n(a+l)  (180). 

The  sign  + is  to  be  prefixed  to  d(n—  1)  when  the  progression  is 
an  increasing  one,  and  — when  decreasing. 

In  these  two  F ormulas  we  have  the  five  quantities,  a , d,  n,  l , s ; hence 
if  any  three  of  these  quantities  be  given,  the  values  of  the  other  two 
may  be  .found  from  the  two  Equations,  (121). 


HARMONIC AL  PROGRESSION. 

(182.)  An  harmonic al  progression  is  a series  of  quantities  such 
that,  of  any  three  consecutive  terms , the  first  : the  third  ::  the  differ- 
- ence  between  the  first  and  second  : the  difference  between  the  second 
and  third. 

Thus  the  numbers  3,  4,  6,  12,  are  in  harmonical  progression, 

since  3 : 6 ::  4 — 3 : 6 — 4, 
and  4 : 12  ::  6 — 4 : 12—6. 

An  Harmonical  Proportion  consists  of  four  terms  such  that  the 
first  is  to  the  fourth  as  the  difference  between  the  first  and  second  is  to 
the  difference  between  the  third  and  fourth. 

Thus  a,  b,  c,  d,  are  in  Harmonical  Proportion,  if 

a : d ::  a—b  : c — d ; 
or  a \ d ::  b — a : d — c. 

The  numbers  16,  8,  3,  2,  are  in  Harmonical  Proportion,  since 
16  : 2 ::  16  — 8 : 3— 2 . 

I - * 

The  first  and  last  terms  of  the  Progression  or  proportion  are  called 
the  two  extremes,  and  all  the  intermediate  terms  the  means. 

a* 


114 


PROGRESSION. 


An  Harmonical  converted  into  an  Arithmetical 
Progression. 

(183.)  The  reciprocals  of  the  terms  of  an  Harmonical  progression, 
are  in  Arithmetical  progression. 

Let  a,  b,  c,  be  three  consecutive  terms  of  a decreasing  harmonica', 
progression. 

Then  a : c ::  a — b : b—c,  (182); 

Converting  this  proportion  into  an  equation  wc  have 
ab — ac—ac—bc. 

Dividing  each  term  in  this  equation  by  abc,  and  reducing  +ha  eer 
eral  quotients  to  their  lowest  terms,  we  find 
1 _ _ _1_  _ 1 
c b b a 

Transposing  the  first  and  the  last  term  of  this  equation 


We  thus  find  that  the  difference  between  the  reciprocals  of  a and  b, 
is  equal  to  the  difference  between  the  reciprocals  of  b and  c ; hence 
these  reciprocals  are  in  Arithmetical  Progression,  (175). 

The  numbers  3,  4,  6,  12,  form  an  Harmonical  Progression,  (182): 
by  taking  the  reciprocals  of  the  several  terms  we  have  the  Arithmeti- 
cal Progression 

b b b iV; 

in  which  the  common  difference  of  the  terms  is  jj. 

(184.)  An  harmonical  mean  between  two  given  terms,  is  equal  to 
twice  their  product  divided  by  their  sum. 

From  the  first  of  the  preceding  equations,  namely, 
ab—ac—ac — be, 

we  shall  find  b = — — ; 

a+c 

and  b is  the  harmonical  mean  between  a and  c. 

The  harmonical  mean  between  3 and  6.  is 
3x6x2  36 

3 + 6 - T ~ ' 


PROGRESSION. 


115 


GEOMETRICAL  PROGRESSION. 

(185.)  A geometrical  progression  is  a series  of  quantities  in 
which  each  succeeding  term  has  the  same  ratio  to  the  term  which 
immediately  precedes  it. 

■ 

I Tims  1,  2,  4,  8,  16,  is  an  increasing  Progression  in  which  each 
succeeding  term  is  double  the  one  which  immediately  precedes  it ; that 
is,  the  ratio  of  the  progression  is  2. 

And  27,  9,  3,  1,  -J-,  is  a decreasing  progression  in  which  each  suc- 
ceeding term  is  one-third  of  the  one  which  immediately  precedes  it ; 
and  the  ratio  of  the  progression  is  consequently 

Hence  the  successive  terms  of  a Geometrical  Progression  consist  of 
the  first  term  multip>lied  continually  into  the  ratio,  that  is,  multiplied 
into  the  successive  powers  of  the  ratio. 

The  theory  of  Geometrical  Progression  is  contained  in  the  following 
propositions. 

The  Last  Term. 

(186.)  The  last  term  of  a Geometrical  Progression,  is  equal  to  the 
first  term  X that  power  of  the  ratio  which  is  expressed  by  the  number 
of  terms  less  one. 

Let  a he  the  first  term,  and  r the  ratio  of  the  progression ; then, 
multiplying  a continually  into  r,  the  series  will  be 

a,  ar,  ar2,  ar3,  ar 4,  &c. 

Since  the  ratio  r begins  in  the  second  term,  with  exponent  1,  its 
! exponent  in  the  last  term  will  always  be  one  less  than  the  number  of 
terms  ; hence  the  last  term  consists  of  the  first  X into  that  power  of  * 
which  is  expressed  by  the  number  of  terms  minus  1. 

1(187  ) Cor.  The  last  term  oi  a Geometrical  Progression  -f-  the  first 
term,  gives  that  power  of  the  ratio  which  is  expressed  by  the  number 
of  terms  less  one. 

Thus  nr4  H-a=r4  ; the  number  of  terms  being yiTe. 

F* 


116 


PROGRESSION. 


Product  of  the  two  Extremes. 

(188.)  The  'product  of  the  two  extremes  in  a Geometrical  Progres 
sion,  is  equal  to  the  product  of  any  two  terms  equidistant  from  them. 
or  to  the  square  of  the  middle  term  when  the  number  of  terms  is  odd. 

Let  a be  the  first  term,  and  r the  ratio  of  the  progression;  then, 
multiplying  a into  the  successive  poivers  of  r,  the  series  is 
a,  ar , ar2,  ar 3,  ar4,  &c. 

Of  these  five  terms  the  product  of  th e first  and  the  last,  is 
axar4=a2r4. 

The  product  of  the  second  and  the  fourth,  which  are  equidistant 
from  the  extremes,  is  arxa?3,  also  =a2r4. 

We  perceive  moreover  that  the  product  of  the  two  extremes  is 
equal  to  the  square  of  the  middle  term  or2. 

In  like  manner  the  proposition  will  he  found  true  for  any  number 
of  terms. 

(189.)  Cor.  A geometrical  mean,  or  a mean  proportional,  between  j 
two  given  terms,  is  equal  to  the  square  root  of  the  product  of  those  1 
terms. 

For  the  product  of  the  two  given  terms,  considered  as  the  two  ex-  ■ 
tremes  of  a Geometrical  Progression,  is  equal  to  the  square  of  the  mean  ! 
or  middle  term. 


The  Sum  of  all  the  Terms. 

(190.)  The  sum  of  all  the  terms  of  a Geometrical  Progression,  is  < 
equal  to  the  difference  between  the  first  term  and  the  product  of  the  j 
last  term  X the  ratio,  — the  difference  between  the  ratio  and  a 
unit. 

Let  S represent  the  sum  of  the  terms,  and  we  shall  have 
S=a-\-ar-\-ar2  -f-ar3  -\-ar4 , See. 

Multiplying  both  sides  of  this  equation  by  the  ratio  r, 

Sr=zar-\-ar2  -f  ar3  -\-ar4  + ar5 . 

Subtracting  the  first  equation  from  the  second,  we  find 
Sr — S=  ar 5 —a  ; 

which  gives  — — . 

° r—  1 

In  the  numerator  of  this  value  of  S,  observe  that  ar5  is  the  last 
term  ar4  of  the  progression  X the  ratio  r.  The  demonstration  will 
evidently  apply  io  any  number  of  terms. 


PROBLEMS. 


117 


(191.)  The  sura  of  an  infinite  number  of  terms  in  a decreasing 
Geometrical  Progression,  is  equal  to  the  first  term  divided  hy  the  dif- 
ference between  the  ratio  and  a unit. 

In  a decreasing  progression  the  terms  continually  diminish  in  a 
constant  ratio  ; and  if  the  number  of  terms  be  infinite , the  last  term 
will  be  0.  The  last  term  X the  ratio  will  then  he  0,  and  the  expres- 
sion for  the  sum  of  the  terms,  found  above,  will  become 


The  divisor  in  this  case  is  1 — r,  because  the  ratio  of  the  progression 
being  a proper  f raction,  is  less  than  a unit. 

Formulas  in  Geometrical  Progression. 

(192.)  In  a Geometrical  Progression,  let  a he  the  first  term,  r the 
ratio,  n the  number  of  terms,  l the  last  term,  and  s the  sum  of  all  the 
terms.  Then 

0 . . . . l=arn~',  (186), 

D . . . .S=1^,  (190). 

When  the  progression  is  a decreasing  one,  and  the  number  ol 
terms  is  infinite, 

E . . . .S=~  , (191). 

1 — r 

In  the  Formulas  C and  D we  have  the  five  quantities,  a,  r,  n,  l,  s ; 
hence  if  any  t/iree  of  these  quantities  be  given,  the  values  of  the  other 
two  may  be  found  from  the  two  Equations,  (121). 

The  principles  which  have  been  established  in  this  Chapter  may 
be  applied  to  the  solution  of  the  following 

Problems  in  Progressions. 

1.  The  first  term  of  an  increasing  Arithmetical  progression  is  3, 
the  common  difference  of  the  terms  is  2,  and  the  number  of  terms  20. 
What  is  the  last  term  ? and  the  sum  of  all  the  terms  ? 

.Ans.  41,  and  440. 

2.  The  first  term  of  a decreasing  Arithmetical  progression  is  100, 
the  common  difference  of  the  terms  is  3,  and  the  number  of  terms  34. 
What  is  the  last  term  ? and  the  sum  of  all  the  terms  ? 

Ans.  1,  and  1717. 


118 


PROBLEMS. 


3.  What  is  the  sum  of  the  numbers  1,  2,  3,  4,  5.  &c.,  continued 

to  1000  terms?  Ans.  500500. 

4.  What  is  the  common  difference  of  the  terms  in  an  Arithmetical 

progression  whose  first  term  is  10,  last  term  150,  and  number  of  terms 
21  ? Ans.  7. 

5.  If  the  third  term  of  an  Arithmetical  progression  be  40,  and  the 

fifth  term  70,  what  will  the  fourth  term  be  ? Ans.  55. 

6.  If  the  first  term  of  an  Arithmetical  progression  be  5,  and  the 
fifth  term  30,  what  will  the  second,  third,  and  fourth  terms  be  ? 

Find  the  common  difference,  (177),  and  thence  the  three  interme- 
diate terms.  Ans.  11^;  17-^-;  23f. 

7.  If  the  fourth  term  of  an  Arithmetical  progression  be  37,  and  the 
eighth  term  60,  what  are  the  intermediate  terms 

Ans.  42f;  48J,  54£. 

8.  What  is  the  sum  of  25  terms  of  an  increasing  Arithmetical  pro- 

gression in  which  the  first  term  is  •£,  and  the  common  difference  of  the 
terms  also  £ ? (176).  Ans.  162£. 

9.  The  first  term  of  an  increasing  Arithmetical  progression,  is  1, 
and  the  number  of  terms  23.  What  must  be  the  common  difference , 
that  the  sum  of  all  the  terms  may  be  100  ? 

Let  x represent  the  common  difference ; 
then  1 + 22*  is  the  last  term,  (176) ; 

2 + 22* 

and - — X 23  is  the  sum  of  the  terms,  (ISO). 

2 

Hence  an  Equation  may  be  formed  from  which  the  value  of  * will  I 
be  found. — Or  we  might  substitute  the  numbers  1,  23,  and  100,  for  a, 
n,  and  s in  Formulas  A and  B,  ( 1 S 1 ),  and  find  the  value  of  d,  as  one 
of  the  two  unknown  quantities.  Ans.  fj. 

10.  If  the  first  term  of  a decreasing  Arithmetical  progression  is 

100,  and  the  number  of  terms  21,  what  must  the  common  difference 
be,  that  the  sum  of  the  series  may  be  1260  ? 4. 

11.  A and  B start  together,  and  travel  in  the  same  direction;  A 

goes  40  miles  per  day  ; B goes  20  miles  the  first  day.  and  increases  Iris 
rate  of  travel  J}  of  a mile  per  day.  How  far  will  they  be  apart  ai  llie  | 
end  of  40  days  1 Ans.  215  miles. 

12.  One  Hundred  stones  being  placed  on  the  ground  in  a straight 

line,  at  the  distance-  of  2 yards  from  each  other ; how  far  will  a person 
travel  who  shall  bring  them,  one  by  one,  to  a basket  which  is  placed  2 
yards  from  the  first  stone?  Ans.  11  miles  840  yards. 


PROBLEMS. 


119 

13.  Find  the  third  term  of  an  Harmonical  progression,  whose  first 
and  second  terms  are  12  and  15  respectively. 

If  x represent  the  third  term;  we  shall  have 

12  : x ::  15  — 12  : a;— 15.  Ans.  20. 

14.  What  is  the  first  term  of  an  Harmonical  progression  whose 

second  and  third  terms  are  30  and  20  respectively?  Ans.  60. 

15.  What  is  the  fourth  term  of  an  Harmonical  proportion  whose 
first,  second,  and  third  terms  are  2,  3,  and  8 respectively  ? 

Ans.  16. 

16.  If  the  first  and  third  terms  of  an  Harmonical  progression  be  25 
and  40  respectively,  what  will  the  second  term  be  ? Ans.  30y|. 

17.  The  first  and  fourth  terms  of  an  Harmonical  progression,  are  10 
i and  20  respectively.  What  are  the  two  intermediate  terms  ? 

This  problem  may  be  solved  by  finding  two  arithmetical  means 
between  yg  and  yy,  and  then  taking  the  reciprocals  of  the  terms  thus 
! found,  (183).  Ans.  12,  and  15. 

18.  The  fifth  and  eighth  terms  of  an  Harmonical  progression  are 
! 20  and  40  respectively.  What  are  the  two  intermediate  terms  ? 

Ans.  24,  and  30. 

19.  The  first  term  of  a Geometrical  progression  is  2,  the  ratio  of 
the  progression  is  3,  and  the  number  of  terms  4.  What  is  the  last 

! term  l and  the  sum  of  all  the  terms?  Ans.  54,  and  80. 

20.  The  first  term  of  a Geometrical  progression  is  f the  ratio  of 
the  progression  is  and  the  number  of  terms  4.  What  is  the  last 

i term  ? and  the  sum  of  all  the  terms  ? Ans.  gy,  and  |y. 

21.  What  is  the  sum  of  an  infinite  number  of  terms  in  the  Geome- 
’ tncal  progression  whose  first  term  is  100,  and  ratio  \ ? Ans.  133-J-. 

22.  What  is  the  sum  of  an  infinite  number  of  terms  in  the  Geome- 
trical progression  'whose  first  term  is  300,  and  ratio  y?  Ans.  450. 

23.  If  the  first  and  third  terms  of  a Geometrical  progression  are  8 
and  72  respectively,  what  is  the  second  term  ? 

- The  second  term  is  equal  to  the  square  root  of  8 X72,  (189). 

1 • Or,  considering  the  third  as  the  last  term  of  the  progression, 

72-f-8  = 9 is  the  square  of  the  ratio , (187)  ; 
then  3 is  the  ratio  of  the  progression ; and  the  second  term  is  now 
readily  obtained.  Ans.  24. 

24.  If  the  third  and  fifth  terms  of  a Geometrical  progression  be  75 
and  300  respectively,  what  will  the  fourth  term  be?  Ans.  150 


120  PROBLEMS. 

25.  If  the  first  and  fourth  terms  of  a Geometrical  progression  are  3 
and  24  respectively,  what  are  the  two  intermediate  terms  ? 

Am.  6 and  12. 

26.  If  the  seventh  and  tenth  terms  of  a Geometrical  progression 
are  6 and  750  respectively,  what  are  the  intermediate  terms? 

Am.  30  and  150. 

27.  What  is  the  sum  of  an  infinite  number  of  terms  in  the  series 
h i,  ■$,  &c.,  in  which  the  ratio  of  the  progression  is  evidently  -f  ? 

Ans.  2. 

28.  If  a body  move  forever  at  the  rate  of  2000  feet  the  first  second 

1000  the  second,  500  the  third,  and  so  on,  what  is  the  utmost  distance 
it  can  reach?  Am.  4000. 

29.  If  10  yards  of  cloth  he  sold  at  the  rate  of  $1  for  the  first  yard 
$2  for  the  second.  $4  for  the  third,  and  so  on,  what  would  he  the  price 
of  the  last  yard  ? and  what  would  the  whole  amount  to  ? 

. Am.  $512,  and  $1023. 

30.  If  13  acres  of  land  were  purchased  at  the  rate  of  $2  for  the 

first  acre,  $6  for  the  second,  $18  for  the  third,  and  so  on,  what  would 
the  last  acre  amount  to?  Am.  $1062882. 

31.  Allowing  the  interest  of  a sum  of  money  to  be  $500  the  first 

year,  $400  the  second,  $320  the  third,  and  so  on,  forever,  what  would 
be  the  whole  amount  of  interest?  Am.  $2500. 

32.  Two  bodies  move  at  the  same  time,  from  the  same  point,  in 

opposite  directions.  One  goes  2 miles  the  first  hour,  4 the  second,  6 
the  third,  and  so  on  ; the  other  goes  2 miles  the  first  hour,  4 the  se- 
cond, 8 the  third,  Ac.  ; how  far  will  they  be  apart  at  the  end  of  12 
hours  ? Ans.  8346  miles. 

33.  A and  B set  out  at  the  same  time  to  meet  each  other.  A tra- 
vels 3,  4,  5,  &c.  miles  on  successive  days,  and  B 3,  4^-,  6f-,  &c.  miles 
on  successive  days.  They  meet  in  10  days;  what  is  the  distance  be- 
tween the  two  places  from  which  they  traveled  ? 

Ans.  414|^  miles. 


121 


CHAPTER  YIII. 

PERMUTATIONS  AND  COMBINATIONS.— INVOLUTION.— BINOMIAL 
THEOREM.— EVOLUTION. 

PERMUTATIONS. 

(193.)  Permutations  are  the  different  orders  of  succession  in  which 
a given  number  of  things  may  he  taken — either  the  whole  number  to- 
getheror  the  whole  number  taken  two  and  two,  or  three  and  three , 
&c. 


Thus  the  different  Permutations  of  the  three  letters  a,  b,  and  c, 
when  all  are  taken  together,  are 

abc,  acb,  bac,  cab,  bca,  cba. 

And  the  different  Permutations  of  the  same  letters  when  taken  two 
and  tivo,  are  ab,  ba,  ac,  ca,  be,  cb. 

Number  of  Permutations. 

(194.)  If  n represent  a given  number  of  things,  the  number  of  'per- 
mutations that  can  be  formed  of  them,  will  be  equal  to 
n(n  — 1 ){n — 2 )(n — 3 )(n — 4),  and  so  on, 
until  the  number  of  factors  multiplied  together  is  equal  to  the  number 
of  things  taken  in  each  permutation. 

To  demonstrate  this  proposition, — suppose  that  we  have  n letters, 
a,  b,  c,  d,  &c.,  to  be  subjected  to  Permutations. 

If  we  reserve  one  of  the  letters,  as  a,  there  will  remain  n — 1 let- 
ters ; and  by  writing  a before  each  of  the  remaining  letters,  we  have 

n—  1 permutations  of  n letters,  taken  tico  and  two,  in  which  a 
stands  first. 

In  like  manner  we  should  find  n—  1 permutations  of  n letters, 
taken  two  and  two,  in  which  b stands  first ; and  so  for  each  of  the  n 
letters.  Hence  we  shall  have 

n[n — 1)  permutations  of  n letters  taken  two  and  two. 

Suppose  now  that  the  n letters  are  to  be  taken  three  and  three. 

By  reserving  a,  and  proceeding  with  n — 1 letters  as  before,  we 
should  find  ( n — 1 )(n  — 2)  permutations  of  n — 1 letters  taken  two  and 
two  ; and  by  writing  a before  each  of  these  permutations,  we  have 


122 


PERMUTATIONS. 


[n—  l)(n  — 2)  permutations-  of  n letters,  taken  three  and  three , in 
which  a stands  first. 

We  should  find  the  same  number  of  permutations  of  n letters, 
taken  three  and  three , in  which  b stands  first ; and  so  for  each  of  the  a 
letters.  Hence  we  shall  have 

n[n — 1 )(n  — 2)  permutations  of  n letters  taken  three  and  three. 

Suppose  now  that  the  n letters  are  to  be  taken  four  antifour. 

By  reserving  a,  and  pursuing  the  operation  in  the  same  manner  as 
before,  we  should  find 

n[?i — 1 )(n  — 2 )(n  — 3)  permutations  of  n letters  taken  four  and 
four. 

Thus  the  demonstration  proceeds  ; the  number  of  factors  multiplied 
together  being  found  always  equal  to  the  number  of  letters  taken  in 
each  permutation. 


As  an  Example  of  the  application  of  the  principle  atove  demon 
strated, — suppose  it  were  required  to  determine  the  number  of  Permu- 
tations, or  di  fferent  orders  of  successio?i,  that  could  be  formed  in  a class 
composed  of  six  pupils,  by  taking  the  whole  number  in  each  permu 
tation. 

Since  the  six  pupils  are  to  be  taken  in  each  Permutation,  the  num- 
ber of  factors  to  be  employed  is  six ; hence  the  number  of  permuta- 
tions is 

6x5x4x3x2x1=720. 

If  the  whole  six  were  to  be  subjected 'to  Permutations  by  taking 
five  at  a time,  the  number  of  permutations  would  be 
6x5x4x3x2  = 720. 

If  the  whole  six  were  to  be  subjected  to  Permutations  by  taking 
four  at  a time,  the  number  of  permutations  would  be 
6x5x4x3  = 360. 

03?=*  It  will  be  observed  above,  that  the  number  of  Permutations 
will  be  the  same,  whether  the  whole  number  of  things,  or  one  less  than 
the  whole  number,  be  taken  in  each  permutation. 


COMBINATIONS. 


123 


COMBINATIONS. 

(195.)  Combinations  are  the  different  collections  which  may  be 
formed  out  of  a given  number  of  things,  by  taking  the  same  number  in 
ea_h  collection — without  regard  to  the  order  of  succession. 

Thus  the  different  Combinations  which  may  be  formed  out  of  the 
three  letters  a , b,  and  c,  by  taking  two  at  a time,  are 

ab,  be,  ac. 

Observe  that  ab  and  ba  are  not  different  combinations,  hut  different 
■permutations,  of  the  letters  a and  b. 

In  Permutations  we  have  regard  to  the  order  of  succession,  and 
may  therefore  have  two  permutations  of  two  things.  In  Combinations 
we  do  not  consider  the  order  of  succession;  so  that  the  combination  of 
two  or  more  things  is  the  same,  in  whatever  order  they  are  taken. 

Number  of  Combinations. 


(196.)  If  n represent  a given  number  of  things,  the  number  of  com- 
binations that  can  be  formed  out  of  them,  will  be  equal  to 
n{n — 1 )(n — 2 )(n- 


1.2. 3. 4 


and  so  on, 


until  the  number  of  factors  in  the  dividend,  and  also  in  the  divisor,  is 
equal  to  the  number  of  things  taken  in  each  combination. 


To  demonstrate  this  proposition,  we  observe  that  the  rmmerator  in 
the  preceding  expression,  is  the  number  of  permutations  of  n things 
taken  four  and  four,  (194). 


On  the  same  principle,  the  denominator  4x3x2xl  is  the  num- 
ber of  permutations  of  four  things  taken  all  together. 

Now  since  there  can  be  hut  one  combination  of  4 things  taken 
all  together,  the  number  of  permutations  of  n things  taken  four  and 
four  is  1.2. 3. 4 times,  that  is,  24  times,  the  number  of  combinations. 


Hence  the  number  of  Combinations  of  n things  taken  four  and 
four,  is  of  the  number  of  Permutations  ; and  will  therefore  be  found 
by  dividing  the  numerator  by  the  denominator. 

This  mode  of  demonstration  may  evidently  be  applied  whatever  be 
the  number  of  things  taken  in  each  combination. 


324 


PROBLEMS. 


PROBLEMS 

In  Permutations  and  Combinations. 

1.  In  how  many  different  ways  might  a company  of  10  person! 

seat  themselves  around  a table  ! (194).  Ans.  3628800. 

2.  How  many  different  numbers  might  be  expressed  by  the  10  nu- 
meral figures,  if  5 figures  he  used  in  each  number!  Ans.  30240. 

3.  In  how  many  different  ways  may  the  names  of  the  12  months 

of  the  year  be  arranged  one  after  another!  Ans.  479001600 

4.  How  many  different  permutations  of  8 men  could  be  formed  out 

of  a company  consisting  of  15  men  1 Ans.  259459200. 

5.  In  how  many  different  ways  might  the  seven  prismatic  colors,  . 

red,  orange,  yellow,  green,  blue,  indigo,  and  violet,  have  been  arranged 
in  the  solar  spectrum  1 Ans.  5040. 

6.  How  many  different  combinations  of  two  colors  could  be  formed 

out  of  the  7 prismatic  colors  ! (195).  Ans.  21. 

7.  How  many  different  combinations  of  5 letters  may  he  formed 

out  of  the  26  letters  of  the  Alphabet!  Ans.  65780. 

8.  How  many  different  combinations  of  2 elements  might  be  formed  I 

out  of  the  56  elements  described  in  Chemistry  ! Ans.  1540. 

9.  In  how  many  different  ways  might  a company  of  20  men  be 
arranged,  in  single  file,  in  a procession  1 

Ans.  2432902008176640000. 

10.  A farmer  wishes  to  select  a team  of  6 horses  out  of  a drove 

containing  10  horses.  How  many  different  choices  for  the  team  will  1 
he  be  able  to  make!  Ans.  210. 

11.  In  how  many  different  ways  might  the  planets  Mercury,  Venus, 

the  Earth,  Mars,  Jupiter,  Saturn,  Uranus,  and  Neptune  succeed  one 
another  in  tlfe  solar  system  ? Ans.  40320. 

12.  A company  of  20  persons  engaged  to  remain  together  so  long  j 
as  they  might  be  able  to  combine  in  different  couples  in  their  eve- 
ning walks.  What  time  will  be  required  to  fulfil  the  engagement ! 

M;iS.  1 90  days. 

13.  How  many  different  permutations  of  7 letters  might  be  formed 

out  of  the  26  letters  of  the  Alphabet  ? Ans.  3315312000. 

14.  In  an  exhibition  of  a Public  School,  5 speakers  are  to  he  taken 
from  a class  of  15  students.  How  many  different  selections  of  the  five 
might  be  made  1 and  in  how  many  different  ways  might  the  5 sue-  1 
ceed  one  another  in  the  delivery  of  their  speeches  1 

Ans.  3003  and  120. 

15.  Out  of  a Company  consisting  of  100  soldiers  Six  are  to  be 
taken  for  a particular  service.  How  many  different  selections  of  the  6 I 
might  be  made  1 and  in  how  many  different  ways  might  the  6 chosen 
be  disposed  with  regard  to  the  order  of  succession  ? 

Ans.  1102052400;  and  720 


INVOLUTION. 


12D 


INVOLUTION. 

(197  ) Involution  consists  in  raising  a given  quantity  to  any  re- 
quired power.  This  may  always  be  effected  by  multiplying  the 
quantity  into  itself  as  many  times  less  one  as  there  are  units  in  the  ex- 
ponent of  the  power. 

Thus  aa  is  a2 , the  second  power  or  square  of  a ; 

aaa  is  a3,  the  third  power  or  cube  of  a ; and  so  on. 

Observe  that  one  multiplication  of  a into  itself  produces  the  second 
power  of  a ; two  multiplications  produce  the  third  power,  and  so  on  ; 
also  that  the  number  of  times  the  quantity  becomes  a factor  in  raising 
a Power,  is  equal  to  the  exponent  of  the  Power. 

(198.)  A higher  poioer  of  a quantity  may  also  be  found  by  multi- 
plying together  two  or  more  lower  powers  (of  the  same  quantity)  the 
| sum  of  whose  exponents  is  equal  to  the  exponent  of  the  rpquired 
power. 

Thus  a 2 Xfl2  produces  a 4 ; a2  x a3  produces  a5,  &c. 

Powers  of  Unity , Monomials,  Fractions,  Spc. 

(199.)  Every  power  of  unity  is  unity , since  any  number  of  Is  mul- 
tiplied together  produce  only  1 ; thus  lxlXl  &c.  =1. 

(200.)  A monomial  is  raised  to  any  required  Power  by  raising  its 
numerical  coeffcie?it  to  that  power,  and  multiplying  the  exponents  of 
its  other  factors  by  the  exponent  of  the  power. 

Thus  to  find  the  third  power  of  4ax2,  we  raise  4 to  its  third  power, 
which  is  4 X 4 X 4 = 64,  and  multiply  the  exponents  of  a and  x by  3 : we 
thus  obtain  64a  3 a:6. 

Observe  that  multiplying  the  given  exponents  hy  the  exponent  of 
the  required  Power,  is  only  a brief  method  of  performing  the  requisite 
multiplications  in  the  Involution  of  Molromials. 

(201.)  A fraction  is  raised  to  any  required  Power,  by  raising  its 
numerator  and  denominator,  separately,  to  that  power. 

d 2 d 6 

Thus  the  third  power  of  — is  - 3-  . (98). 

| 

A mixed  quantity  may  be  raised  to  any  required  Power  by  involv- 
ing the  equivalent  improper  fraction. 


126 


INVOLUTION. 


The  Sign  to  be  Prefixed  to  a Power. 

i (202.)  Every  even  'power  of  a quantity  is  positive  ; while  every  odd 
power  has  the  same  sign  as  the  quantity  f rom  which  it  is  derived. 

Thus  if  a he  positive , all  its  powers,  as  aa,  aaa,  and  so  on,  will 
evidently  he  positive  ; hut  if  a be  negative  we  shall  have 
— a. — a=a2  ; a2. — a— — a3;  — a3. — a = a4  ; a 4. — a— — a5,  &c. 
from  which  it  is  plain  that  all  the  even  powers,  as  the  2d,  4th,  and  so 
on,  will  be  positive , while  all  the  odd  powers  will  be  negative. 

EXERCISES. 


On  the  Powers  of  Monomials. 


1. 

Find  the  square  of  3 ax2. 

Ans.  9a2*4 

2. 

Find  the  cube  of  — 2 a2x. 

Ans.  — 8a6*3. 

3. 

Find  the  square  of  — 4a*2. 

Ans.  16a2*4. 

4. 

Find  the  cube  of  3a3*. 

Ans.  27a9x3. 

5. 

Find  the  square  of  ab2c3. 

Ans.  a2b*c 6. 

6. 

Find  the  cube  of  —a2x2y. 

Ans.  — a6x6y3. 

7. 

Find  the  square  of  \ax2 . 

Ans.  ^a2*4. 

8. 

Find  the  cube  of  \ay3 . 

Ans.  £ a3y 9. 

9. 

Find  the  square  of  — | -abn. 

Ans.  %a2b2n. 

10. 

Find  the  cube  of  — |ri*2. 

Ans.  — fpz3x6. 

11. 

Find  the  4th  power  of  2a". 

. Ans.  16a4n. 

12. 

Find  the  4th  power  of  — \a2. 

Ans.  jL as. 

13. 

Find  the  4th  power  of  — 3*2. 

Ans.  81*8. 

14. 

Find  the  5th  power  of  \x2 . 

.Ans.  ^2*10- 

15. 

Find  the  6th  power  of  — ax2. 

Ans.  a6*12 

16. 

Find  the  6th  power  of  2 y2 . 

Ans.  64»/12 

17. 

Find  the  7th  power  of  — a2yn. 

Ans.  — a14iy"'*. 

18. 

Find  the  7th  power  of  \y. 

Ans. 

19. 

Find  the  8th  power  of  a 2 bn. 

Ans.  a16is". 

20. 

Find  the  8th  power  of  — %x2. 

jins,  ale*16 

BINOMIAL  THEOREM. 


127 


Powers  of  Polynomials. 

(203.)  The  Powers  of  a binomial , or  of  any  polynomial , may  be 
obtained  by  successive  multiplications  of  the  quantity  into  itself,  (197). 

Thus  the  2d  power  or  square  of  a-\-b,  is 

(a-\-b){a-\-b)=za2  -\-2ab-\-b2 . 

And  the  3d  power  or  cube  of  a-\-h,  is 

(a+6)(a-f  Z>)(a  + d)  = a3  + 3a25  + 3a62+63. 

The  Involution  of  Binomials,  however,  and  thence  of  Polynomials, 
is  greatly  facilitated  by  the  application  of  Newton’s 

Binomial  Theorem. 

(204.)  This  Theorem  explains  a general  method  of  developing  a 
Binomial  according  to  any  exponent  with  which  the  Binomial  may  be 
affected. 

In  developing  ( a-\-b )2  by  the  Binomial  Theorem  we  should  obtain 
the  2d.  power  or  square  of  ( a-\-b ). 

JL 

In  developing  (a-^b)'1  we  should  obtain  the  square  root  of  (a-\-b) ; 
and  so  for  other  exponents. 

Development  of  {a±b)n  ; 
n representing  any  exponent. 

1.  The  Exponent  of  a in  the  first  term  of  the  development,  is  the 
same  as  the  exponent  of  the  Binomial,  and  decreases  by  the  subtrac 
tion  of  1,  continually,  in  the  succeeding  terms. 

2.  The  Exponent  of  b entering  as  & factor,  commences  with  1 in 
the  second  term  of  the  development,  and  increases  by  the  addition  of  1, 
continually,  in  the  succeeding  terms. 

3.  The  Coefficient  of  the  first  term  is  1 ; that  of  the  second  term  is 
the  same  as  the  exponent  of  the  Binomial  ■ — and  the  coefficient  of  the 
second,  or  of  any  term,  X the  exponent  of  a in  that  term,  and  -h  by 
the  exponent  of  b increased  by  1,  gives  the  coefficient  of  the  succeeding 

term. 

4.  The  Signs  of  the  terms  in  the  development  of  (affb)  will  all 
be  + ; while  for  ( a — b)  they  will  be  alternately  + and  — . 

From  these  principles  we  have  the  following 


128 


BINOMIAL  THEOREM. 


(205.)  Formula  for  the  Development  of  the  Binomial  (a  + 5)". 


(a4 -b)n=an-\-nan  16  + 


*(*-!)  _—**»  , n(n-l)(n-2)  n 
2 2><3  “ 


'353,  icc 


When  the  exponent  n is  a positive  integer , as  2,  3,  or  4,  &c.,  the 
development  will  terminate  at  the  term  in  which  the  exponent  of  b 
becomes  equal  to  the  exponent  of  the  Binomial. 

For  the  exponent  of  a in  that  term  will  be  0,  and  0 will  thus  be- 
come a factor  in  finding  the  coefficient  of  the  next  term ; hence  the 
uext  term  will  be  0,  (43). 


example  I. 


To  find  the  4th  power  of  («  + 5),  by  the  Binomial  Theorem  ; 
that  is,  to  develop  (a  + 6)4. 

The  literal  factors  without  the  coefficients,  will  be 
a4  a3b  a2b 2 ab3  b 4 

By  computing  and  introducing  the  coefficients,  with  the  sig>is,  we 
have  a4  + 4a3§+6a262+4a63  + 64. 


The  student  will  readily  perceive  the  application  of  the  preceding 
principles  to  the  several  terms  of  this  Power. 

Demonstration  of  the  Binomial  Theorem. 

The  principles  which  have  been  given  for  the  development  of 
(aFb)n,  will  be  demonstrated  under  the  supposition  that  the  exponent 
n is  a positive  integer. — A general  demonstration  would  be  equally 
applicable  to  negative  or  fractional  exponents  ; such  demonstration  is 
unnecessary  here,  and  is  too  abstruse  for  the  present  stage  of  our  sub- 
ject. 

If  we  multiply  together  the  binomial  factors 
a-\-b,  a+c,  a+d,  a-\-e, 

and  decompose  the  product  terms  which  contain  the  lower  powers  of  a, 
the  final  Product  may  be  represented  as  follows ; 


+6 

a3  -\-bc 

a 2 -\-bcd 

+ C 

-\-bd 

-\-hce 

-\-d 

-\-be 

-\-bde 

-\-cd 

-\-cde 

-j-ce 

-\-de 

-\-bcde 


BINOMIAL  THEOREM. 


129 


In  this  Product  observe  that  the  coefficients  of  the  powers  of  a in 
the  successive  terms,  are  as  follows ; 

The  Coefficient  of  a4  is  1 ; the  coefficient  of  a 3 is  the  sum  of  the 
second  terms,  b,  c,  See.,  of  the  binomial  factors  composing  the  Product ; 
the  coefficient  of  a 2 is  the  sum  of  the  products  be,  bd,  See.,  of  the  second 
terms  of  the  binomial  factors  combined  two  and  two,  (196) ; and  the 
coefficient  of  a is  the  sum  of  the  products  bed,  bee,  See.,  of  the  second 
terms  of  the  binomial  factors  combined  three  and  three. 

Suppose  now  that  c,  d,  e,  & c.  are  each  equal  to  b,  and  the  number 
of  binomials  equal  to  n.  The  Product  of  these  factors  will  be  the  wth 
power;  of  (a -\-b) ; that  is,  it  will  be 

the  development  of  («+£)”. 

The  Exponent  of  a in  the  first  term  of  the  Product  will  evidently 
be  n ; and,  as  exemplified  in  the  preceding  multiplication,  this  expo- 
nent will  decrease  by  1,  continually,  in  the  succeeding  terms. 

Hence  we  shall  have 

an,  a ”~1,  an  2,  See.  in  the  consecutive  terms. 

The  Coefficient  of  an  will  evidently  be  1.  The  coefficient  of  a"-1 
i in  the  second  term,  will  be  n times  b. 

Hence  the  second  term  will  be 

nban~'i  or  nan~xb. 


The  Coefficient  of  a™"  2 in  the  third  term,  will  be  b 2 taken  as  many 
times  as  there  are  combinations  of  two  letters  in  n letters,  (196). 

Hence  the  third  term  will  be 


n{n — 1) 
1X2 


b2an  2 


n{n—  1) 
or  — - — — ^ an 
1x2 


-2b2. 


The  Coefficient  of  a”-3  in  the  fourth  term,  will  be  b3  taken  as 
many  times  as  there  are  combinations  of  three  letters  in  n letters. 
Hence  the  fourth  term  will  be 


n(n-l)(n-2) 

1x2x3 


or 


n(n — l)(n—  2) 

1x2x3 


3b3. 


The  several  terms  thus  obtained  agree  with  Formula  (205) ; 

I and  the  law  of  development  thus  indicated,  will,  in  like  manner,  be 
found  applicable  to  any  number  of  terms. 

With  regard  to  the  Signs, — if  is  evident  that  when  a and  b are 
positive,  all  the  terms  in  any  power  of  (a + 5)  will  be  positive,  (42). 

When  b is  negative,  all  the  odd  powers  of  b will  be  negative, 
(202) ; and  these  negative  powers  multiplied  by  the  positive  powers  of 
a,  will  cause  the  2d,  4th,  6th,  Sec.  terms  to  become  negative. 

7 


BINOMIAL  THEOREM. 


rjo 

EXAMPLE  II. 

To  find  the  5th  power  of  the  binomial  a — x. 

The  literal' factors  without  the  coefficients,  are 

as,  aix,  a3%2,  a2z 3,  ax*,  x3 . 

By  computing  and  inserting  the  coefficients,  with  the  signs,  we 
have 

a 5 — 5a4x+  10a3x2  — 10a2x3-f-5a:z4 — x5 . 


EXERCISES 

On  the  Powers  of  Polynomials. 

1.  Find  the  square  of  a — x.  Ans.  a2 — 2 aP-\-x2. 

2.  Find  the  cube  of  a-\-y.  Ans.  a3  p3a2yp3ay2  -\-y3 . 

3.  Find  the  cube  of  ap2b. 

By  applying  the  principles  of  the  Binomial  Theorem,  we  obtain 
a3  P3a2 .2bp3a{2b)2  P(2b)3  ; 

which  may  be  developed  into  Ans.  a3 p8a2bp\2ab2  p8b3 . 

4.  Find  the  square  of  3«  + y.  Ans.  9a2  p6aypy2 . 

5.  Find  the  cube  of  2a  — 3x.  Ans.  8 a3 — 36ffl2x-f54<zx2 — 27x3. 

6.  Find  the  square  of  a + x — y, 

By  operating  on  (x — y)  as  if  it  were  a monomial,  and  applying  the 
Binomial  Theorem  to  ap(x — y ),  wre  obtain 

a2-\-2a(x — y)+(z— y)2  ; 

which  may  be  developed  into  A}is.  a2 + 2ax—2ay+x2 — 2xyPy2 . 

7.  Find  the  square  of  a — bpy.  Ans.  a2 —2ab-\-2ay-\-b2 — 2 bypy2. 

8.  Find  the  square  of  a — x — y.  A?is.  a2 — 2ax+z2 — 2ayp2xyPy2- 

9.  Find  the  square  of  apb  — 2x. 

Ans.  a2  p2abpb2 — 4ax — 4Sx+4x2. 

10.  Find  the  square  of  a2px3. 

This  square,  according  to  the  Binomial  Theorem,  may  be  indicated 
thus  : 

(a2)2  + 2a2x3  + (x3)2  ; 

which  may  be  developed  into  Ans.  a4+2a2x3+ar6 

11  Find  the  cube  of  a2 — y3 . Ans.  a 6 — 3a*y3p3a2yG — ?/9 

12.  Find  the  cube  of  1 +2x2.  Ans.  l+6x2  + 12x4  + Sx6. 

13  Find  the  fourth  power  of  d-|-x. 

Ans.  a4+4a3.x+6a2x2  + 4ax_3+x4 
14.  Find  the  fifth  power  of  a — y. 

A™-'  w 5 — 5a*yP10a3y2  — 10 a2 y3 p5ay*  — y 5 


EVOLUTION. 


131 


EVOLUTION. 

(200.)  Evolution  consists  in  extracting  any  required  root  of  a given 
quantity,  regarded  as  the  corresponding  power  of  the  root  to  be  found. 

Extracting  the  square  root  consists  in  finding  a quantity  whose 
square  is  equal  to  a given  quantity  ; extracting  the  cube  root  consists 
in  finding  a quantity  whose  cube  is  equal  to  a given  quantity  ; and  so 


Roots  of  Unity , Monomials , Fractions,  Spc. 

(207.)  Every  root  of  unity  is  unity , since  the  square,  or  cube,  &c. 
of  1 is  1 ; thus  1x1  = 1;  1 X 1 X 1 = 1 ; and  so  on. 

In  general  terms,  any  poivcr  or  root  of  a unit  is  a unit. 

(208.)  Any  required  Root  of  a monomial  will  he  found  by  extract- 
ing the  root  of  its  numerical  coefficient,  and  dividing  the  exponents  of 
its  literal  factors  by  the  integer  corresponding  to  the  root. 

• 

Thus  the  square  root  of  25  a2x  is  5ax^,  found  by  extracting  the 
square  root  of  25,  and  dividing  the  exponents  of  a and  x by  2. 

i . 1 

And  the  cube  root  of  27 a6x2  is  3a2x6,  found  by  extracting  the 
cube  root  of  27,  and  dividing  the  exponents  of  a and  x by  3. 

The  correctness  of  this  method  will  appear  from  considering  that 
the  Extracting  of  a Root  is  the  reverse  of  raising  the  corresponding 
Power,  (200). 

Hence  also  the  propriety  of  denoting  roots  by  fractional  expo- 
nents. 

i i i 

a 2 denotes  the  square  root  of  a,  because  aF  Xr —a. 

(209.)  Any  required  Root  of  a fraction  will  be  found  by  extracting 
he  root  of  its  numerator  and  denominator,  separately. 

Thus  the  square  root  of  is  - — ^ , found  by  extracting  the 

*)cl  y 6 a y 

quare  root  of  the.  numerator,  and  the  square  root  of  the  denominator. 

A Root  of  a mixed  quantity  would  be  found  by  extracting  the 
not  of  the  equivalent  improper  f raction. 

I G " 


132 


EVOLUTION. 


(210.)  A Root  whose  exponent  is  resolvable  into  two  factors,  may 
be  found  by  extracting  in  succession  the  roots  denoted  by  those  factors. 

The  4th  root  may  be  found  by  extracting  the  square  root  of  the 
iquare  root, — the  exponent  \ which  denotes  the  4th  root,  being  x ■£. 

Thus  the  4th  root  of  x*  is  the  square  root  of  x 2 ; the  4th  root  of 
81  is  the  square  root  of  9;  and  the  4th  root  of  10000  is  the  square 
root  of  100. 

The  6th  root  may  he  found  by  extracting  the  square  root  of  the 
cube  root , or  the  cube  root  of  the  square  root, — being  equal  to  • 

The  correctness  of  this  method  of  extracting  Roots,  is  evident  from 
considering,  that,  by  raising  in  succession  the  'powers  denoted  by  two 
or  more  factors,  we  shall  obtain  the  power  denoted  by  the  product  o£ 
those  factors. 

Thus  the  square  of  the  cube  of  a is  (a3)2=a6,  (200),  and  there- 
fore, conversely,  the  square  root  of  the  cube  root  is  the  sixth  root 

Roots  of  Pouters  or  Powers  of  Roots. 

(211.)  The  Numerator  of  a fractional  exponent  denotes  a power  of 
the  quantity  affected,  and  the  Denominator  a root  of  that  power. 

Thus  a 2 denotes  the  square  root  of  the  first  power  of  a,  or  simply 

2 I 

the  square  root  of  a.  In  like  manner  a:i  denotes  the  cube  root  of  the 

3. 

square  of  a ; a4  denotes  the  4th  root  of  the  cube  of  a ; and  so  on. 
But,  observe  that 

(212. )■  A Root  of  any  power  of  a quantity,  is  equal  to  the  same 
power  of  the  same  root  of  the  quantity. 

To  illustrate  this  principle  it  may  he  shown  that  the  cube  root  of 
the  6th  poiver  of  a,  is  equal  to  the  6th  power  of  the  cube  root  of  a. 

The  cube  root  of  a6  is  a2  since  a2.a2  .a2  =a6  ; and  the  6th  power 
of  a 3 is  also  a2,  since  the  cube  of  a's  is  a,  and  the  square  of  this  cube, 
— which  gives  the  6th  power  of  a:J, — is  a2,  (200). 

To  give  an. application  of  this  principle  to  numbers, — the  cube  root 
of  the  square  of  8,  is  4,  and  the  square  of  the  cube  root  of  S is  4 


EVOLUTION. 


133 


Equivalent  Exponents. 

(213.)  Two  Exponents  which  are  numerically  equivalent,  are  also 
equivalent  exponentially ; and  may  therefore  be  substituted,  the  one 
for  the  other. 

2 1 

Thus  a4  =a2  ; the  fourth  root  of  a2  is  equivalent  to  the  square 

root  of  a.  The  4th  root  of  a2  is  a-,  since  a2  raised  to  the  4th  power 
produces  a2. 

3.  1 4.  2 4. 

In  like  manner  x6~x2  ; y6=y3  ; a2 —a2  ; &c. 

To  give  an  example  in  numbers , — the  fourth  root  of  the  square  of 
9,  that  is,  of  81,  is  3,  and  the  square  root  of  9 is  also  3.  . 

On  this  principle  a fractional  exponent  may  always  be  taken  in  its 
1 lowest  terms. 

The  Sign  to  be  Prefixed  to  a Root. 

(214.)  Every  odd  Root  of  a quantity  has  the  same  sign  as  the 
quantity  itself. 

For  the  quantity  itself  is  an  odd  power  with  reference  to  such  root, 
(206),  and  an  odd  power  Has  the  same  sign  as  the  quantity  of  which  it 
is  a power,  (202). 

(215.)  Every  even  Root  of  a positive  quantity  is  ambiguous , that 
is,  the  quantity  itself  does  not  determine  whether  the  root  is  positive  or 
negative. 

This  follows  from  the  principle  that  every  even  power  of  a nega- 
tive, as  well  as  of  a positive  quantity,  is  positive,  (202). 

The  square  of  — a,  as  well  as  of  +a,  is  -\-a2  ; hence  the  square 

{root  of  a2  is  +a  or  else  — a ; and  when»it  is  not  known  whether  a2 
was  derived  from  a or  — a,  it  is  uncertain  which  of  these  two  is  the 
square  root. 

This  uncertainty  as  to  the  sign  of  the  root,  is  expressed  by  the  am- 
biguous sign  ± plus  or  minus  ; thus 

the  square  root  of  a2  is  ±a. 

For  a like  reason  the  fourth  root  of  a4  is  ±<z. 

The  square  root  of  9 is  ±3  ; the  fourth  root  of  16  is  ±2  ; &c. 

(216.)  An  even  Root  of  a negative  quantity  is  impossible,  since 
there  is  no  quantity  which  can  be  raised  to  an  evenriegative  power, 
(202). 


134 


EVOLUTION. 


Thus  no  quantity  multiplied  into  itself  will  produce  — a2  ; this 
quantity  therefore  has  no  square  root.  For  a like  reason  — a4  has  no  j 
square  root  ox  fourth  root. 

In  like  manner  — 9 has  no  square  root ; — 16  has  no  square  root  or 
fourth  root ; — 25a2  has  no  square  root,  &c. 

EXERCISE  S 


On  the  Koois  of  Monomials. 


1. 

Find  the  square  root  of  4a2 x4. 

Ans.  ±2ax2. 

2. 

Find  the  cube  root  of  8 a3y6. 

Ans.  2ay 2. 

3. 

Find  the  square  root  of  9 a4x. 

Ans.  ±3 a2xf. 

4. 

Find  the  cube  root  of  — 27 y2. 

2 

Ans.  — 3 y3. 

5. 

i 

Find  the  square  root  of  16a;2. 

A?is.  ±4x4. 

6. 

Find  the  cube  root  of  a3xy 2. 

1 2 

Ans.  axfyi. 

7. 

Find  the  square  root  of  25a3. 

_3 

Ans.  zb  5a2. 

8. 

Find  the  cube  root  of  — 64y2. 

2 

Ans.  — 4y3 

9. 

Find  the  square  root  of  36a;6. 

Ans.  ±6a:3. 

10. 

Find  the  cube  root  of  8 ayn. 

J.  J> 

Ans.  2 a3y3. 

11 

Find  the  square  root  of  -\a4y2. 

Ans.  ±ha2y. 

12 

Find  the  cube  root  of  — 8a3x3n. 

Ans.  — 2 ax*. 

13. 

Find  the  square  root  of  $jcy4. 

A7is.  d=f.rTy2. 

14. 

Find  the  cube  root  of  — -£aa:3. 

j. 

Ans.  —\a3x. 

15. 

Find  the  fourth  root  of  16a4y2. 

al;zs.  ±2 ay2. 

16. 

Find  the  cube  root  of  — fjx2. 

•1 

Ans.  — 

17. 

Find  the  fourth  root  of  pt-ay2 . 

1 1 

Ans.  d=  4a4yh 

18. 

Find  the  fifth  root  of  — a10a:2y. 

2 1 

Ans.  — a2xsy5. 

19. 

Find  the  sixth  root  of  a3x2y. 

1 4 I 
±a'-x3y6. 

20 

* 

Find  the  sixth  root  of  ha2y3n. 

1 1 > 
Ans.  z tb6a''yi' 

EVOLUTION. 


135 


Roots  of  Polynomials. 

(217.)  The  method  of  extracting  any  required  Root  of  a Polynomial, 
may  be  discovered  from  the  manner  in  which  the  corresponding  power 
of  a polynomial  is  formed.  This  subject  will  be  elucidated  under  the 
appropriate  Rules. 

RULE  XVIII. 

(218.)  To  Extract  the  Square  Root  of  a Polynomial. 

. 1.  Arrange  the  Polynomial  according  to  the  powers  of  one  of  its 
letters,  and  take  the  square  root  of  the  left  hand  term , for  the  first 
term  of  the  root. 

2.  Subtract  the  square  of  the  root  thus  found  from  the  given  Poly- 
nomial ; divide  the  remainder  by  twice  the  root  already  found,  and 
annex  the  quotient  to  both  the  root  and  the  divisor. 

3.  Multiply  the  divisor  thus  formed  by  the  last  term  in  the  root ; 
subtract  the  product  from  the  dividend  ; divide  the  remainder  by  twice 
the  root  now  found  ; and  so  on,  as  before. 

EXAMPLE. 

To  extract  the  Square  Root  of 

a2  Jr2ab-\-b2 . 

The  required  Root,  we  already  know,  is  a-\-b,  since  the  square  of 
this  binomial  is  the  given  trinomial : — our  object  is  to  show  that  the 
root  a + 5 would  be  found  by  the  foregoing  Rule. 

a 2 + 2ab + b 2 (a + b 
a 2 

2 a-\-b)  2ab  + b2 

2 ab  + b2  ■' 

The  first  term  a of  the  root,  is  the  square  root  of  a 2 , the  left  hand 
term  of  the  given  polynomial.  Subtracting  the.  square  of  a , we  have 
the  remainder  2 ab  + b2. 

We  have  now  to  discover  a divisor  of  this  remainder,  which  will 
give,  for  a quotient,  the  next  term  b of  the  root. 

2a,  that  is,  twice  the  root  already  found,  divided  into  2 ab,  gives  b ; 
and  b annexed  to  2 a makes  the  divisor  2 a-{-b.  This  divisor  multiplied 
by  b,  produces  2ab-\-b2 , — which  completes  the  operation. 

The  Rule  is  framed  in  accordance  with  the  process  thus  discovered. 


136 


EVOLUTION. 


EXERCISES 


On  the  Square  Root  of  Polynomials. 

1.  Find  the  square  root  of  the  polynomial 

a4 — 4a3+6a2 — 4a+l-  Ans.  a2  —2a+  1. 

2.  Find  the  square  root  of  the  polynomial 

a4-j-4a3z-h6a2x2  -\-4axz -{-x* . Ans.  a2  + 2ax+z2 . 


3.  Find  the  square  root  of  the  polynomial 

1 — §y-\-\2>y2  — 12y3jr4y*.  Ans.  l—3y-i-2y2 

4.  Find  the  square  root  of  the  polynomial 

4x6 — 4a:4  + 12a;3  + a:2 — 6x+9.  Ans.  2x3 — ar+3. 


5.  Find  the  square  root  of  the  polynomial 

4a4  + 12a3x+  13a2x2  ~i~6ax3 +x4.  A?is.  2a2  + 3aa:+a;2. 


6.  Find  the  square  root  of  the  polynomial 

1 — 6a+13a2  — 12a3-f-4a4.  Ans.  1— 3<z+2a2. 

7.  Find  the  square  root  of  the  polynomial 

a;4  + 2aa:3  + 3a2a;2  +2a3a:  + a4.  Ans.  x2 -\-ax-\-a2 . 

8.  Find  the  square  root  of  the  polynomial 

9a4  — 12a37/+28a2i/2  — 16a?/3  + 16?/4.  Ans.  3a 2 — 2ay-\-4y2. 

9.  Find  the  square  root  of  the  polynomial 

25  + 34a:2  + 12a;3  + 20a;+9a;4u  Ans.  5 + 2x+3x2. 

10.  Find  the  square  root  of  the  polynomial 

a4  -\-4a3y~ l2ay-\-4a2y2 + § — 6a2.  Atis.  a2-\-2ay  — 3. 

11.  Find  the  square  root  of  the  polynomial 

4 + 24a;2 — 16a:+4a:4 — 16a:3.  Ans.  2 — 4a:+2:r2. 

12.  Find  the  square  root  of  the  polynomial 

1—  2y  + 7y2—  2y3 +5y4 + 12y5 +4y6 . Ans.  1 — y+3y2  +2y3. 


EVOLUTION. 


137 


Square  Root  of  Numbers. 

• 

The  preceding  Rule  is  substantially  the  same  as  the  Arithmetical 
Rule  for  extracting  the  square  root  of  Numbers.  To  show  its  applica- 
tion, however*  to  the  latter  purpose,  we  must  premise  the  following 
principles. 

(219.)  If  a Number  he  separated  into  periods  of  two  figures  each, — 
from  right  to  left, — these  periods  will  correspond,  respectively,  to  the 
units , tens,  hundreds,  &c.,  in  the  Square  Root  of  the  number. 

For  since  the  square  of  ten  is  100,  the  square  of  the  tens  figure  in 
the  root  will  leave  tivo  vacant  places  in  the  right  of  the  given  number ; 
these  two  places  must  therefore  correspond  to  the  units  in  the  root. 

And  since  the  square  of  a hundred  is  10000,  the  square  of  the 
hundreds  in  the  root  leaves  four  vacant  places  in  the  right  of  the 
number ; and  the  first  two  corresponding  to  the  'units,  the  next  two 
must  correspond  to  the  tens  in  the  root. 

In  like  manner  it  is  shown  that  the  third  period  of  two  figures 
corresponds  to  the  hundreds  in  the  square  root  of  the  given  number  ; 
and  so  on. 

(220.)  If  a Number  be  divided  into  any  two  parts,  the  Square  of 
the  number  will  be  equal  to  the  square  of  the  first  part  + twice  the 
first  X the  second  + the  square  of  the  second  part. 

For,  a representing  the  first  part  of  the  number,  and  b the  second , 
a-\-b  will  he  equal  to  the  number ; and 

( a-\-b )2  —a2  -\-2ab-\-b2 . 

With  these  principles  established  we  may  proceed  to  the  following 
examples. 

1.  To  extract  the  Square  Root  of  529 
5’29  ( 23 

4 

43)  129 
129 

The  first  period  29  corresponds  to  the  units  figure,  and  the  5 
to  the  tens  figure  in  the  root,  (219). 

The  greatest  integral  square  root  of  the  5 is  2 tens , the  square  of 
which  is  4 hundreds,  and  this  subtracted  leaves  129. 

7* 


13S 


EVOLUTION. 


Since  the  root  of  the  given  number  consists  of  two  parts,  tens  and 
units , and  the  square  of  the  2 tens  has  been  subtracted,  the  remainder 
129  = twice  2 tens  X the  units  + the  units2 , (220). 

Doubling  the  root,  2 tens,  already  found,  we  take  the  product,  4 
tens,  for  a divisor.  - The  0 in  4 tens,  or  40,  may  he  omitted  in  finding 
the  next  figure  in  the  root,  if  we  omit  the  9 in  the  corresponding  place 
in  the  129. 

We  therefore  say  4 in  12,  3 times.  This  3 annexed  to  the  4 
makes  the  4 now  become  4 tens.  Then  43,  or  4 tens  + 3 units,  X 3 
units,  produces  the  remainder  129  ; and  thus  completes  the  operation. 

2.  To  extract  the  Square  Root  of  36  84  49 
36’84’49  ( 607 
36 

1207  ) 84  49 

84  49 

Pointing  off  periods  of  two  figures,  from  right  to  left,  we  find  that 
"he  root  will  contain  units,  tens,  and  hundreds,  (219). 

The  square  root  of  the  left  hand  period  36,  is  6,  the  square  of 
which  subtracted  leaves  no  remainder  in  that  period. 

We  take  the  next  period  84  to  find  the  next  figure  in  the  root. 
Doubling  the  root  6,  for  a divisor,  and  omitting  the  4,  we  say  12  in  8,  j 
0 time. 

Including  the  next  period  49,  and  doubling  the  root  60  now  found, 
for  a divisor,  we  say  120  in  844,  7 times. 

Annexing  7 to  120,  the  divisor  becomes  1207,  which  multiplied  by 
7 produces  8449 ; and  thus  the  operation  is  completed. 

Square  Root  of  Decimals. 

(221.)  In  extracting  the  Square  Root  of  a Decimal  Fraction,  the 
■periods  must  he  taken  from  the  decimal  point  towards  the  right ; and  ! 
a 0 must  be  annexed,  if  necessary,  to  complete  the  last  period. 

The  last  period  must  be  complete,  because,  by  the  principles  of 
decimal  multiplication,  the  square  of  a decimal  Fraction  must  contain 
tivice  as  many  decimal  figures  as  are  in  the  root. 

The  number  of  decimal  figures  to  he  made  in  the  root,  is  therefore 
the  same  as  the  number  of  decimal  periods. 

When  an  exact  root  camiot  be  found,  decimal  periods  of  00  each 
may  be  annexed,  and  the  root  continued  in  decimals  to  any  required 
exactness. 


EVOLUTION.  139 

EXERCISES 

On  the  Square  Root  of  Numbers. 

1.  Find  the  square  root  of  11236,  and  of  §|-|. 

Ans.  106  ; and 

2.  Find  the  square  root  of  38809,  and  of  . 

Ans.  197  ; and 

3.  Find  the  square  root  of  75076,  and  of  | ^-§-1 . 

Ans.  274  ; and  |J. 

4.  Find  the  square  root  of  22801,  and  of  fflf. 

Ans.  151  ; and 

5.  Find  the  square  root  of  .582  169,  and  of  127J|. 

Reduce  the  to  an  eqivalent  decimal,  before  proceeding  with  the 
evolution. 

Ans.  .763  ; and  11.292'. 

6.  Find  the  square  root  of  475.125,  and  of  346^J. 

Ans.  21.797’;  and  18.616’. 

7.  Find  the  square  root  of  37.4780,  and  of  470j^g. 

Ans.  6.121’;  and  21.684’ 


RULE  XIX. 

(222.)  To  Extract  the  cube  Root  of  a Polynomial. 

1.  Arrange  the  Polynomial  according  to  the  powers  of  one  of  its 
letters,  and  take  the  cube  root  of  the  left  hand  term,  for  the  first  term 
of  the  root. 

2.  Subtract  the  cube  of  the  root  thus  found  from  the  given  Polyno- 
mial, and  divide  the  remainder  by  3 times  the  square  of  the  root  al- 
ready found,  and  annex  the  quotieyit  to  the  root. 

3.  Complete  the  divisor  by  adding  to  it  the  product  which  arises 
from  annexing  the  last  term  in  the  root  to  3 times  the  other  part  of  the 
root,  and  midtiplying  the  result  by  the  last  term. 

4.  Multiply  the  completed  divisor  by  the  last  term  in  the  root,  and 
subtract  the  product  from  the  dividend. 

5.  Find  the  next  incomplete  divisor  by  adding  to  the  last  complete 
divisor  doe  product  which  completed  it,  and  the  square  of  the  last  term 
in  the  root ; divide,  and  complete  the  divisor,  as  before ; and  so  on. 

G* 


140 


EVOLUTION. 


EXAMPLE. 

To  extract  the  Cube  Root  of 

a3  + 3a26+3ad2  + &3. 

The  required  Root,  we  already  know,  is  a+d,  (203),  and  our  ob- 
ject is  to  show  that  this  root  would  be  found  by  the  Rule. 

a3  + 3a26  + 3aZ>2  + Z>3  ( a+5 
a 3 

3a2  + 3a6+52  ) oa2b-\-'3ab2  -\-b3 

3a26+3a62+d3 


The  first  term  a of  the  root,  is  the  cube  root  of  a3,  the  left  hand 
term  of  the  given  polynomial.  Subtracting  a3,  we  have 
the  remainder  3a26  + 3ad2 +63. 

We  have  now  to  find  a divisor  of  this  remainder,  which  will  give, 
for  a quotient,  the  next  term  b of  the  root. 

3a2,  that  is,  3 times  the  square  of  the  root  already  found,  divided 
into  3 a2b,  gives  b.  Adding  3 ab,  and  also  b2 , to  3a2,  the  completed 
divisor  is  3a2+3a6  + b2. 

This  divisor  multiplied  by  b,  produces  the  last  dividend,  and  thus 
shows  that  the  operation  is  completed. 


EXERCISES 


On  the  Cube  Root  of  Polynomials. 


1.  Find  the  cube  root  of  the  polynomial 

a6  — 6a5  + 15a4  — 20a3  + 15a2  — 6a+l.  Ans.  a2— 2a+l. 

2.  Find  the  cube  root  of  the  polynomial 

1 — 6x+21x2 — 44x3  + 6ox4 — 54xs  + 27x6.  Ans.  1 — 2x+3x2. 

3.  Find  the  cube  root  of  the  polynomial 

8 — 12x+30x2— 25x3  + 30x4  — 12x5  + 8x6.  Ans.  2— x+2x2. 


4.  Find  the  cube  root  of  the  polynomial 

8 + 36?/2  + 24iy+32?/3  + 6ys+?/6  + 18?/4.  Ans.  2+2 ypy2 

5.  Find  the  cube  root  of  the  polynomial 

9a4 — 3a5 — 13a3+a6  — 12a  + 8+ 18a2.  Ans.  a2 — a + 2. 

6.  Find  the  cube  root  of  the  polynomial 

21x4— 44x3  — 54x  + 63x2 — 6x5+x6  + 27.  Ans.  x2 — 2x+3. 

7.  Find  the  cube  root  of  the  polynomial 

y6  — 9«/5+39i/4—  99?/3  + 156^2  — 144i/+64.  A/iS.  y2—3y-rl 


EVOLUTION. 


141 


Cube  Root  of  Numbers. 

The  method  of  extracting  the  Cube  Root  of  Numbers  involves  the 
following  principles. 

(223.)  If  a Number  be  separated  into  "periods  of  three  figures  each, 
— from  right  to  left, — these  periods  will  correspond,  respectively,  to  the 
units,  tens,  hundreds , &c.,  in  the  Cube  Root  of  the  number. 

For  since  the  cube  of  ten  is  1000,  the  cube  of  the  tens  figure  in 
the  root,  will  leave  three  vacant  places  in  the  right  of  the  given  num- 
ber ; these  three  places  must  therefore  correspond  to  the  units  in  the 
root. 

And  since  the  cube  of  a hundred  is  1000000,  the  cube  of  the  hun- 
dreds in  the  root  leaves  six  vacant  places  in  the  right  of  the  number  ; 
apd  the  first  three  corresponding  to  the  units , the  next  three  must  cor- 
respond to  the  tens  in  the  root. 

In  like  manner  it  is  shown  that  the  third  period  of  three  figures 
corresponds  to  the  hundreds  in  the  cube  root  of  the  given  number,  and 
so  on. 

(224.)  If  a Number  be  divided  into  any  two  parts,  the  Cube  of  the 
number  will  be  equal  to  the  cube  of  the  first  part  + 3 times  the  square 
of  the  first  X the  second  + 3 times  the  first  X the  square  of  the 
second  + the  cube  of  the  second. 

For  a representing  the  first  part  of  the  number,  and  b the  second, 
a-\-b  will  be  equal  to  the  number  ; and 

(a + 6) 3 =a3  + 3<z26+3a&2  +£3. 

As  an  example  of  the  application  of  this  principle  to  numbers,  let 
275  be  divided  into  the  two  parts  200  and  75 ; then 

(200  + 75)3:=:2003  + 3x2002  X 75  + 3x200  x752  + 753. 

If  the  values  of  the  several  terms  on  the  right  be  computed,  and 
added  together,  we  shall  obtain  the  cube  of  275. 

We  might  now  proceed  to  extract  the  Cube  Root  of  Numbers  in  a 
manner  corresponding  to  the  Rule  already  given  for  the  cube  root  of 
Polynomials  ; but  the  method  by  the  following  Rule  is  more  direct  and 
simple. 


142 


EVOLUTION. 


RULE  XX. 

(225.)  To  Extract  the  Cube  Root  of  a Number. 

1.  Sep_rate  the  number  into  'period s of  three  figures  each,  from 
right  to  len,  observing  that  the  last  period  will  sometimes  have  but 
one  or  two  figures. 

2.  Take  the  greatest  integral  cube  root  of  the  left  hand  period,  for 
the  first  figure  of  the  root  required, — subtract  its  cube  from  said  period, 
— and  to  the  remainder  affix  the  next  period  for  a dividend,. 

3.  Take  3 times  the  square  of  the  root  already  found,  for  an  in- 
complete divisor  ; divide  it  into  the  dividend  exclusive  of  its  two  right 
hand  figures,  and  annex  the  quotient  to  the  root. 

4.  Complete  the  divisor  by  annexing  to  it  00,  and  adding  the  pro- 
duct which  arises  from  annexing  the  last  figure  in  the  root  to  3 times 
the  other  part  of  the  root,  and  multiplying  the  result  bij  the  last 
figure. 

5.  Multiply  the  completed  divisor  by  the  last  figure  in  the  root; 
subtract  the  product  from  the  dividend  ; and  to  the  remainder  affix  the 
next  period  for  a new  dividend. 

6.  Find  the  next  incomplete  divisor  by  adding  to  the  last  complete 
divisor  the  product  which  completed  it,  and  the  square  of  the  last 
figure  in  the  root ; divide,  and  complete  the  divisor,  as  before ; and  so 
on. 

In  applying  this  Rule  it  will  be  convenient  to  have  the  following 

Table  of  Roots  and  Cubes. 

Roots,  1, 2, 3, 4, 5, 6, 7 8, 9. 

Cubes,  1, 8,  . . . 27,  . . 64,  . . 125,  . . 216,  . . 343,  . . 512, . . 729. 

We  will  first  apply,  and  then  demonstrate  this  Rule. 


EXAMPLE. 

To  extract  the  Cube  Root  of  95  44  39  93. 


95’44  3’9  93 

64 

48 

31  44  3 

4800  + 625  = 54  25 

27  12  5 

5425+  625+  25  = 60  75 

43  1 8 993 

607500+  9499=  6169  99 

43  1 8 993 

£ VOLUTION 


143 


The  greatest  integral  cube  root  of  the  left  hand  period'  95,  is  4,  the 
cube  of  which  subtracted  leaves  the  remainder  3 1 . Affixing  the  next 
period,  we  obtain 

the  dividend  31443. 

Three  times  the  square  of  4,  the  root  already  found,  is  48.  "We 
divide  48  into  314,  excluding  the  43.  The  quotient  would  appear  to 
be  6 ; but  the  divisor  being  as  yet  too  small , we  take  5 for  the  quo- 
tient. 

To  complete  the  divisor , we  annex  00  to  it,  and  add  625  ; — the 
625  being  obtained  by  annexing  the  5 in  the  root  to  3 times  the  4,  and 
multiplying  the  result  125  by  5. 

Multiplying  the  completed  divisor  5425  by  5,  subtracting,  and  af 
fixing  the  next  period  to  the  remainder,  we  find 
the  dividend  43  18  99  3. 

For  the  next  incomplete  divisor,  we  add  to  the  last  complete  divi- 
sor the  product  625  which  completed  it,  and  the  square  25  of  the  5 in 
the  root.  Dividing  by  6075,  the  quotient  is  7. 

To  complete  this  divisor,  we  annex  00,  and  add  9499, — this  last 
number  being  obtained  by  annexing  the  7 to  3 times  the  45,  and  mul- 
tiplying the  result  1357  by  7.  The  divisor  now  found  multiplied  by  7 
produces  the  last  dividend. 


Demonstration  of  the  Rule  for  the  Cube  Root 
of  Numbers. 


To  facilitate  the  Demonstration,  we  will  take  a number  containing 
but  two  periods, — whose  root  will  therefore  consist  of  tens  and  units, 
(223). 

91*125  ( 45 
64 


48 

4800  + 625  =5125 


27125 

127125 


The  4 in  the  root,  found  according  to  the  Rule,  we  know  is  4 tens, 
(223) ; and  the  second  figure  in  the  root  will  be  units. 

Let  t represent  the  tens,  and  u the  units  of  the  root,  then 
the  given  number  —C  -\-2>t2u-{-otu2  +w3,  (224). 

Hence  the  cube  64, — which  is  64  thousand, — of  the  4 tens,  sub- 
tracted, leaves  27125  = 3£2w+3£w2+2<3. 

By  the  Rule  for  the  cube  root  of  a Polynomial,  the  incomplete  divi- 
sor of  this  remainder — to  be  used  for  finding  the  next  figure  or  term  u 
of  the  root,  is  3£2=3  times  (-4  tens)2  =4800. 


144 


EVOLUTION. 


In  dividing  we  omit  the  00,  and  at  the  same  time  exclude  the  25 
in  the  corresponding  places  ef  the  dividend  27125. 

By  the  Buie  just  referred  to,  the  quantity  to  he  added  to  complete 
the  divisor  3 t2,  is 

(3t+w)w=(3x4  tens  +5)x5  = 625  ; 

that  is,  the  product  which  arises  from  annexing  the  last  figure  in  the 
root  to  3 times  the  other  part  of  the  root,  and  multiplying  the  result 
hy  the  last  figure. 

Suppose  now  that  the  given  number  contains  three  periods , and  its 
root,  consequently,  three  figures. 

Begarding  the  two  figures  already  found  as  constituting  the  first 
part  of  the  root,  the  next  incomplete  divisor,  hy  Buie  XIX,  would  be 

I 'St2+{3t-\-u)u-\-(2>t-\-u)u-\-u 2 ; 

which  consists,  first,  of  the  last  complete  divisor  3Z2  + (37+m)m;  secondly, 
of  the  product  (3 t-\-u)u  which  was  added  to  3 12  to  complete  that  divi- 
sor ; thirdly,  of  the  square  of  the  last  figure  u in  the  root. 

In  like  manner  the  Buie  may  he  shown  to  be  applicable,  whatever 
be  the  number  of  figures  in  the  required  root. 


Cube  Root  of  Decimals. 

(226.)  In  extracting  the  Cube  Boot  of  a Decimal  Fraction,  the  pe 
riods  must  be  taken  from  the  decimal  point  towards  the  right,  and  0 
or  00  must  be  annexed,  if  necessary,  to  complete  the  last  period. 

The  last  period  must  be  complete,  because,  by  the  principles  of  de- 
cimal multiplication,  the  cube  of  a decimal  Fraction  must  contain  3 
times  as  many  decimal  figures  as  are  in  the  root. 

The  number  of  decimal  figures  to  be  made  in  the  root,  is  therefore 
the  same  as  the  number  of  decimal  periods. 

When  an  exact  root  cannot  be  found,  decimal  periods  of  000  each 
may  be  annexed,  and  the  root  continued  in  decimals  to  any  required 
exactness. 


EVOLUTION. 


145 


EXERCISES 

On  the  Cube  Root  of  Numbers. 

1.  Find  the  cube  root  of  262144,  and  of  . 

Ans.  64;  and 

2.  Find  the  cube  root  of  2406104,  and  of 

Ans.  134;  and 

3.  Find  the  cube  root  of  22906304,  and  of  -££\XT2- 

Ans.  284  ; and 

4.  Find  the  cube  root  of  479.2735,  and  of 

Ans.  7.825’;  and  20.309’. 

5.  Find  the  cube  root  of  5371.3745,  and  of  3059^g-. 

Ans.  17.513’  ; and  14.51’ 

6.  Find  the  cube  root  of  403.73331,  and  of  .71200^^-. 

Ans.  7.390’  ; and  .892’. 

7.  Find  the  cube  root  of  4370.666,  and  of 

Ans.  16.34’  ; and 

8.  Find  the  cube  root  of  20796875,  and  of  3511^4j£. 

Ans.  275  ; and  15.20. 

9.  Find  the  cube  root  of  .202262003,  and  of  f-g-f-f. 

Ans.  .587  ; and  A-|-. 

10.  Find  the  cube  root  of  103823,  and  of  2460-|. 

Ans.  47  ; and  13.5. 
Extraction  of  the  nth.  Root. 

(227.)  Any  Root  whatever  of  a Polynomial  might  he  extracted, — 
by  taking  the  root  of  its  left  hand  term, — with  this  root  forming  an  in- 
complete divisor, — with  the  quotient  term,  and  the  root  already  found, 
completing  the  divisor, — and  so  on,  in  a manner  depending  on  the 
order  of  the  root  to  be  extracted. 

But  this  method,  which  is  preferable  for  the  square  and  cube , be- 
comes too  complicated  when  applied  to  the  higher  roots. 

By  dispensing  with  the  completed  divisors,  the  operation  may  be 
simplified,  and  the  process  of  Evolution  generalized,  as  follows. 

[ 


146 


EVOLUTION. 


RULE  XXL 

(228.)  To  Extract  any  Root  of  a Polynomial. 

1 Uunderstanding  the  order  of  the  root  to  he  denoted  by  n, — ar- 
range the  Polynomial  according  to  the  powers  of  one  of  its  letters,  and 
take  the  nth  root  of  its  left  hand  term,  tor  the  first  term  of  the  root. 

2.  Subtract  the  nth  power  of  the  root  found  from  the  given  Poly- 
nomial ; and  divide  the  remainder  by  n times  the  (n — 1)  power  of 
this  root,  for  the  second  term  of  the  root. 

3.  Subtract  the  nth  power  of  the  root  now  found  from  the  given  Po- 
lynomial, and,  using  the  same  divisor  as  before,  proceed  in  the  same 
manner  till  the  nth  power  of  the  root  becomes  equal  to  the  given  Poly- 
nomial. 

This  Rule  may  also  he  applied  to  Numbers,  by  taking  n figures  in 
each  period,  from  right  to  left,  for  integers,  and  from  the  decimal  point 
towards  the  right,  for  decimals;  and  there  will  be  less  liability  to 
error  in  finding  the  quotient  figure , if  new  divisors  be  found  for  the 
second  and  subsequent  remainders. 

EXAMPLE. 

To  extract  the  4th  root  of  30  49  800  625. 

30’4980’0625  ( 235 
24=  16 

4x23=32  ) 144980  ....  Exclude  980  in  dividing. 

23 4 = 279841 

4x233=48668  ) 251390625  . . Exclude  625  in  dividing. 

235  4 = 304980 0 6 25 

•The  preceding  root  might  also  be  found  by  two  extractions  of  the 
square  root,  (210) ; thus 

The  square  root  of  3049800625  is  55225  ; and  the  square  root  of 
the  latter  number  is  235. 


147 


CHAPTER  IX. 

IRRATIONAL  OR  SURD  QUANTITIES.— IMAGINARY  QUANTITIES 

Perfect  and  Imperfect  Powers. 

(229.)  A Perfect  Power  of  any  degree,  is  a quantity  which  has  an 
exact  root  of  the  same  degree  ; — otherwise,  the  quantity  is  called  an 
Imperfect  Power. 

Thus  4 and  9a2  are  perfect  squares,  having  the  exact  square  roots 
2 and  3 a ; while  2 a and  8a;3  are  imperfect  squares,  since  they  have 
no  exact  square  roots. 

In  like  manner  8 and  27a3  are  perfect  cubes;  while  9 and  25a:2 
are  imperfect  cubes,  since  they  have  no  exact  cube  roots. 

The  Polynomial  a2-\-2ab-\-b2  is  a perfect  square,  whose  root  is 
a-\-b ; (218);  and  a3 — oa2xp3ax 2 — a:3  is  a perfect  cube,  whose 
root  is  a — x.  • 


Irrational  or  Surd  Quantities. 

(230.)  A Rational  quantity  is  one  which  can  be  accurately  express- 
ed without  any  indicated  root ; as  2,  3a,  or  -fa;. 

An  Irrational  or  Surd  quantity  is  one  which  can  be  accurately 
expressed  only  under  the  form  of  a root, — being  the  indicated  root  ol 
an  imperfect  power. 


i 

Thus  2 2 is  an  irrational  quantity,  since  the  exact  square  root  of  2 
cannot  he  determined.  Also  a * and  a;^  are  irrational  quantities. 


By  the  common  Rule  for  the  square  root  of  numbers,  we  should 
find  the  square  root  of  2=1.414213’  ; hut  other  decimal  figures  would 
succeed  without  end. 


The  term  irrational  when  applied  to  a quantity,  implies  that  such 
quantity  has  no  determinable  ratio  to  unity. 

Irrational  or  Surd  quantities — being  expressed  under  the  form  of 
mots — are  also  called  Radical  quantities,  from  the  Latin  Radix,  a 
root 


148 


IRRATIONAL  OR  SURD  QUANTITIES. 


Radical  Sign. 


(231.)  The  ladical  sign  y/  prefixed  to  a quantity,  denotes  the 
square  root  of  the  quantity ; and  is  therefore  equivalent  to  the  expo- 
nent -J. 

Thus  -J  a is  the  square  root  of  a ; equivalent  to  a2. 


With  the  index  3 affixed  to  it,  this  sign  denotes  the  cube  root 
and  is  then  equivalent  to  the  exponent  -J.  With  the  index  4,  it  de 
notes  the  4th  root ; and  so  on. 


Thus  y/a  is  equivalent  to  a ? ; and  t/ a to  a4. 


The  radical  sign  may  always  he  thus  superseded  by  a fractional 
exponent;  and  this  should  be  done  whenever  any  obscurity  wouli 
arise  in  calculation  from  using  this  sign. 

O C ^ 

For  example,  in  extracting  the  square  root  of  y a,  we  substitute 
the  exponent  and  find  the  required  root  to  be 

a&-~2=a^;  (208). 

A quantity  preceding«the  y/,  without  + or  — interposed,  is  a co- 
efficient or  multiplier  of  the  Surd  ; and  when  no  coefficient  is  express- 
ed, a unit  is  understood. 

. ' Thus  5 V a is  5 times  the  square  root  of  a ; and  V ax  is  the  same 
as  1 V ax. 

Similar  and  Dissimilar  Surds. 


(232.)  Similar  Surds  are  such  as  express  the  same  root  of  the 
same  quantity  ; otherwise,  the  Surds  are  dissimilar. 

Thus  f 3 and  5y/3  are  similar ; while  y/ 3 and  5 "\/3  are  dissimi 

lar.  So  (a  + 6)“  and  oV a -{-b  are  similar  Surds. 

A Rational  Quantity  under  the  Form  of  a Surd. 


(233.)  A rational  quantity  may  be  expressed  under  the  form  of  the 
square  root , or  cube  ro.pt,  &c.,  by  placing  the  corresponding  power  of 
the  quantity  under  the  exponent  or  sign  of  the  root. 

To  express  3 a under  the  form  of  the  square  root , we  pla  ‘.e  the 
square  of  3 a,  which  is  9a2,  under  the  exponent  or  sign  of  the  square 
root,  thus 

3a=(9a2)^  or  V$a2. 


IRRATIONAL  OR  SURD  QUANTITIES. 


149 


Transformation  or  Reduction  of  Surds. 


Certain  Transformations  of  radical  quantities  are  sometimes  neces- 
sary, to  adapt  them  to  the  purposes  of  calculation.  These  depend 
chiefly  on  the  two  following  principles. 


(234.)  When  two  or  more  Factors  have  the  same  exponent,  that 
exponent  may  be  transferred  to  the  product  of  those  factors  ; and, 

Conversely,  the  exponent  of  a Product  may  he  transferred  to  each 
of  the  factors  which  compose  that  product. 


Thus  a1 2 * x2  is  equal  to  (ax)2  ; the  product  of  the  squares  of  a and 
x,  is  equal  to  the  square  of  the  product  of  a and  x. 

1 i i 

And  a2x^  is  equal  to  (ax)2  ; the  product  of  the  square  roots  of  a 
and  x,  is  equal  to  the  square  root  of  the  product  of  a and  x. 

These  principles  result  from  the  methods  of  finding  the  powers  and 
roots  of  quantities  ; thus  hy  squaring  ax, — and  extracting  the  square 
root  of  ax, — we  have 

(az)2  =a2x2,  (200) ; and  (ax)'2  —afx2,  (208). 


These  methods  of  illustration  may  he  applied  to  any  other  power  or 
root  as  well  as  to  the  square  and  square  root.  Thus  the  product  of 
the  cube  roots  of  two  or  more  factors  is  equal  to  the  cube  root  of  the 
product  of  those  factors  ; and  so  for  any  power  or  root. 

To  give  examples  in  numbers  ; — 

VlX  V25=Vl00  ; or2x5  = 10. 
y/8x  1/27^-^216;  or  2x3=6. 

(235.)  The  Exponent  of  a quantity  may  he  changed  to  any  equiva- 
lent expression,  without  altering  the  value  of  the  power  or  root  de- 
noted. 


This  proposition  is  substantially  the  same  as  one  already  demon- 
strated, (213). 


As  examples, 


1 2.  3.  4. 

a 2 =cj4  —a8 —a8  ; 


1 2.  1 

4^=44  = 164=2. 


The  following  are  the  principal  Transformations  required  in  the 

calculation  of  Irrational  or  Surd  quantities. 


L£)0 


IRRATIONAL  OR  SURD  QUANTITIES. 


Product  of  a Rational  and  an  Irrational  Factor. 

(236.)  A rational  multiplier  or  coefficient  may  be  put  under  the 
form  of  the  Surd  annexed,  and  the  product  of  the  two  radicals  be  then 
affected  with  the  same  fractional  exponent  or  radical  sign. 

Thus  aV‘i=V a2  oa2 , by  putting  the  coefficient  a un- 

der the  form  of  the  square  root , (233),  and  transferring  the  sign  -f, 
which  is  equivalent  to  the  exponent  \ , from  the  two  factors  to  the  pro- 
duct 3a2,  (234). 

In  like  manner  2 y/5  = y/4  X = V^IO. 

As  an  example  of  the  utility  of  this  Transformation, — suppose  it 
were  required  to  find  an  approximate  value  of  2y/o. 

By  the  common  method  of  extracting  the  square  root,  we  might 
find  an  approximate  value  of  V 5,  and  such  value  multiplied  by  2 
would  be  an  approximate  value  of  2y/ 5. 

It  is  evident,  however,  that  this  process  would  double,  in  the  pro- 
duct, the  deficiency  in  the  value  found  of  y/ 5 ; and  the  error  thus 
arising  will  be  greater  as  the  coefficient  of  the  Surd  is  greater. 

By  taking  V2Q,  instead  of  2\/5,  the  approximate  value  will  be 
found  independently  of  the  preceding  source  of  error. 

Surds  Reduced  to  Simpler  Forms. 

(237.)  A Surd  is  simplified  by  resolving  the  quantity  under  the 
-y/,  or  exponent,  into  two  factors — one  of  which  is  a perfect  power  \ 
of  like  degree — and  putting  the  extracted  root  of  this  factor  as  a coeffi-  - 
cient  to  the  indicated  root  of  the  other. 

Thus  V/50  = 'v/25x -v/2  = 5y/2; 

or,  50^=  25^ X 2^=5  X 2^,  (234). 

In  dike  manner,  V 4a2+8«3  — V 4 a2V \-\-2a  = 2a\  l-\-2a. 

In  these  examples  the  given  Surds  are  square  roots,  and  the  factors 
25  and  4a2  are  accordingly  perfect  squares. 

As  an  example  of  the  cube  root,  we  have 

y 54=  y/ 27  X y/ 2 = 3\/ 2 ; 27  being  a perfect  cube. 

When  the  given  Surd  is  preceded  by  a Coefficient,  this  coefficient 
must  be  multiplied  into  the  one  found  by  the  process  of  simplifying. 

Thus  3Y^^~^VTGadV2a—oXiaV2a  = l2aV2a. 


IRRATIONAL  OR  SURD  QUANTITIES. 


151 


If,  in  the  preceding  process,  the  square  or  cube  factor  be  the  great- 
est square  or  cube  thus  entering  into  the  composition  of  the  quantity 
under  the  ff , the  Surd  will  be  reduced  to  its  simplest  form. 

Thus  in  each  of  the  examples  above  given,  the  Surd  is  reduced  to 
its  simplest  form.  The  second  example  would  also  give 

V 4a2 + 8 a3  = V 4 V a2  -f-  2a3  =2V a2  + 2a3  ; 
which  is  not  the  simplest  form,  since  a2  is  still  under  the  . 

(238.)  A Fractional  Surd  may  be  reduced  to  an  integral  surd, 
by  multiplying  both  its  terms,  if  necessary,  to  make  the  denominator  a 
perfect  power , and  then  resolving  into  factors,  and  proceeding  as  be- 
fore, (237). 

For  example,  = Vie  = VIS  ^6=  I ^ : 

We  multiply  both  terms  of  the  first  Fraction  by  2,  to  make  the  de- 
nominator 16  a perfect  square.  The  reciprocal  of  this  denominator  is 
the  square  factor,  and  the  numerator  6 is  the  other  factor. 

Surds  which  are  apparently  dissimilar,  often  become  similar 
when  reduced  to  their  simplest  forms.  They  are  thus  prepared  for 
Addition  or  Subtraction,  as  will  be  seen  hereafter. 

Surds  of  Different  Roots  reduced  to  the  Same  Root. 

(239.)  Two  or  more  Surds  of  different  roots,  may  be  reduced  to 
equivalent  ones  of  the  same  root, — by  reducing  their  fraotional  expo- 
nents to  a common  denominator , — raising  each  quantity  under  the 
to  the  power  denoted  by  the  numerator  of  its  new  exponent, — and  ta- 
king the  root  denoted  by  the  common  denominator. 

Thus  to  reduce  y'S  and  y/d  to  the  same  root. 

Reducing  the  exponents  ^ and  | to  a common  denominator, 
we  find  y/5  = 5%,  and  ^4  = 4^,  (235). 

Cubing  the  5,  and  squaring  the  4,  according  to  the  numerators  of 
their  new  exponents,  (211),  we  find 

i ,,  i 
/ 5 = 1256,  and  V4  = 166. 

The  square  root  and  the  cube  root  have  thus  been  both  reduced  to 
the  sixth  root. — This  kind  of  reduction  is  sometimes  necessary  in  the 
Multiplication  and  Division  of  Surds, — as  will  be  seen  hereafter. 


152 


EXERCISES. 


EXERCISES 

On  the  Transformation  or  Reduction  of  Surds. 


1.  Find  the  Product  of  3 -f  A.a.  (236). 

2.  Find  the  Product  of  2y[3x. 

3 Find  the  Product  of  4 -fab. 

4.  Find  the  Product  of  a\f^x: 

5.  Find  the  Product  of  3x^7 . 

6.  Find  the  Product  of  ax^f  10. 

7.  Find  the  Product  of  7 a -\- 1. 

8 Find  the  Product  of  2%/ 1 — x. 

9 Find  the  Product  of  ( a-\-b)f2 . 

10  Find  the  Product  of  (a — x)\/3. 

11  Find  the  Product  of  (a+ 1)%/ x. 

12.  Find  the  Product  of  ( a — 1 )\/y. 


Atis.  y/36a 
Ans.  \f~2Ax. 
Ans.  -\/l6 ab. 
Ans.  oa? x. 
Ans.  V 63z2 
Ans.  V lOa3^3- 
Ans.  -\/49a+49. 

Ans.  ^/S  — tx. 
Ans.  -y/2 (a  + 6)2.  ■ 
Ans.  \/3(a—x)3’ 
Ans. 

Ans.  \fy[a— l)3 


13.  Reduce  -f 24  5ay&  to  its  simplest  form.  (237). 

To  discover  the  greatest  square  factor  of  245,  we  will  divide  this 
number  successively  by  the  square  numbers, 

4,  9,  16,  25,  36,  49,  64,  &c., 

and  take  the  largest  divisor  that  leaves  no  remainder. 


Such  divisor  will  be  found  to  be  49  ; — 19)245(5. 
Then  -f 24 5ays  = -\/49 y*  f 5ay~7y2  \/ day. 


14.  Reduce  y/4 a2b  to  its  simplest  form. 

15.  Reduce  3'fa3x  to  its  simplest  form. 

16.  Reduce  \/8y  to  its  simplest  form. 

17.  Reduce  2\/8 ax2  to  its* simplest  form. 

18.  Reduce  \/27a 4 to  its  simplest  form. 

. 19  Reduce  5y/48 a3y  to -its  simplest  form. 
20.  Reduce  2y/64a4  to  its  simplest  form. 


Ans.  2 a-fb. 
Ans.  3a 'fax 
A>is.  2 \/y 
Ans  \xf2a 
Ans.  3 a\/ a. 
Ans.  20a-y/3ay. 
.Aws.  8a\/a.  ' 


EXERCISES. 

21.  Reduce  3-\/l28x3  to  its  simplest  form. 

22.  Reduce  ey/250 y to  its  simplest  form. 

23.  Reduce  2x-f  432  to  its  simplest  form. 

24.  Reduce  3y\/ 13b  to  its  simplest  form. 

25.  Reduce  \/8-{-l2a3  to  its  simplest  form. 

26.  Reduce  [a-\-b)\/8 lx3  to  its  simplest  form. 

t 

27.  Reduce  3 a2x — 2a3  to  its  simplest  form. 

28.  Reduce  (a — x)\/l92 a3x  to  its  simplest  form. 

29.  Reduce  4-f 4x2  + 8x3  to  its  simplest  form- 


153 

Ans.  24x\/2x 
Ans.  ba\/2y 
Ans.  24x \/ 3 . 
Ans.  9? f\/5. 
Ans.  2^2+3 a3. 
Ans.  3 x(a-\-b) 


Ans.  3a-\f  x— 2a. 
Ans.  4a{a  — x)  \/3x. 
Ans.  8x-fl-{-2x. 


30 


/27 

Reduce  3 W ^ to  an  i?tfegral  Surd  in  its  simplest  form.  (238). 


V- 


31  Reduce  2 V 25  to  an  integral  Surd  in  its  simplest  form. 


32  Reduce  4 


4ax3 

ir 


Ans.  — -/H 

Q 


to  an  integral  Surd  in  its  simplest  form. 


33.  Reduce  a2 


V 3y3 


34.  Reduce  5x  \J  ^ 

v 36 


Qx . 

Ans.  — -yfWax. 
11  v 

to  an  integral  Surd  in  its  simplest  form. 

. . 2a2  3 , - — 

Ans.  — — -v/  18a. 

% 

to  an  integral  Surd  in  its  simplest  form. 

bx 


Ans.  — -f  1 -\-y2. 


2 f „ 

35  Reduce  — v to  an  integral  Surd  in  its  simplest  form. 

5 v a+x 


18 


Ans-  5(5+^  -‘/2(»+4 


8 


104 


EXERCISES 


36.  Reduce  5^/2  and  3\/4  to  Surds  of  the  same  root.  (239). 

By  reducing  the  exponents  of  the  Surd  factor  to  the  common  deno- 
minator, we  find 

13*1.  1 2.  1_ 

2^=26=8e  ; and  4*=46  = 166. 

Ans.  5^/8,  and  3 y/ 16. 

Prefixing  the  rational  co-efficients,  we  have 

5^2=5%/$,  and  3V4  = 3 Vl6. 

37.  Reduce  2y/ 5 and  5y/3  to  Surds  of  the  same  root. 

Ans.  2^/25,  and5y27. 

38.  Reduce  a\/ 5 and  x\/2  to  Surds  of  the  same  root. 

Ans.  a\/ 125,  and  x^/4. 

39  Reduce  1 0 y/ 1 0 and  2^/'3x  to  Surds  of  the  same  root. 

Ans.  lOy/lOO,  and  2^32 

40.  Reduce  7 ^/3 y and  2 y\/  xy2  to  Surds  of  the  same  root- 

Ans.  71'v/81iy4,  and  2yl^/x3y6 

41.  Reduce  a2x  \/\  and  3 y/ a2x  to  Surds  of  the  same  root. 

Ans.  a2x\/\,  and  J \/aix2. 

42.  Reduce  — •(/  2 and  5y\J — to  Surds  of  the  same  root. 

%x  £ oc 

Ans ■ Yx  ^4’  and  5y  * 

43.  Reduce  2 xy/ a and-^yv^  to  Surds  of  the  same  root. 


Ans.  2x\/ a~ , and4;  yb 

y 


44.  Reduce  k y/lO  and  31  v7-  to  Surds  of  the  same  root. 


Ans.  |\/  100,  and  31^/-3' 

45.  Reduce  77-^/J-andv-f/  — to  Surds  of  the  same  root. 

2 yv  a 2x*  y2 

• A*m.  ~ 1 2/It, and  -1-  l2/±! 

2y  V (J4  2*  vys 


ADDITION  AND  SUBTRACTION  OF  SURDS. 


155 


ADDITION  AND  SUBTRACTION  OF  SURDS. 

(240.)  1.  Tlie  Sum,  or  Difference,  of  similar  surds  is  obtained  by 
prefixing  the  sum,  or  difference,  of  their  coefficients  as  a coefficient  to 
the  common  radical  factor. 

2.  Dissimilar  surds  can  be  added  together,  or  subtracted  the  one 
from  the  other,  only  by  the  proper  sign;  but  Surds  apparently  dis- 
similar often  become  similar  when  reduced  to  their  simplest  forms,  (237). 

EXAMPLE. 

To  Add  together  Sy/SOa  and  of  125  a. 

Reducing  the  surds  to  their  simplest  forms,  we  find 
5f80a=5f  16\/Sa=20-\/5a ; 
and  of  \25a=.3f25f  5a=\5f 5a. 

The  two  given  Surds  have  thus  become  similar , (232.)  Adding 
together  the  coefficients  20  and  15,  we  find  the  Sum  o5f  5a. 

_ 

IThe  Difference  of  the  two  given  Surds,  is  (20  — 15)  f 5a  — 5 f 5a. 
The  Addition  or  Subtraction  of  similar  Surds  is  evidently  nothing 
more  than  the  addition  or  subtraction  of  similar  monomials  ; thus 

20  times  f 5a+ 15  times  f 5a  is  35  times  f5a  ; 
just  as  20a+15a  is  35a. 


EXERCISES 

On  the  Addition  and  Subtraction  of  Surds. 


1.  Find  the  Sum  of  3f27  and  2f  48, 

2.  Find  the  Difference  between  f 50  and  fl2. 

3.  Find  the  Sum  of  if  28  and  6-^63. 

4.  Find  the  Difference  between  2\f2  and  5f  18. 

5.  Find  the  Sum  of  -y/ 180  and  ^405. 

H 


Am.  17  f3. 

Arts,  f 2. 
Ans.  32  f 7. 
Ans.  6f2. 
Ans.  \5f5. 


156 


ADDITION  AND  SUBTRACTION  OF  SURDS. 


6.  Find  the  Difference  between -\/l  8 and  2 -\/50.  Ans.  7^2. 

7.  Find  the  Sum  of  -\/ 12 a2  and  -\/2ria 2.  Ans.  5aV 3. 

8.  Find  the  Difference  between  3-\/24x2  and  \/54 x2.  Ans.  3z\/6 

9.  Find  the  Sum  of  4^/3 a and  -\/48 a.  Ans.  8\^3 a 

10  Find  the  Difference  between  V" 4 a3  and  -\/ 9 a3.  Ans.  ay* a. 
11.  Find  the  Sum  of  3 t/4  and  7^/4.  Ans.  10-^4 

12  Find  the  Difference  between  9-\/200  and  -\/288.  Ans.  18^2. 

13  Find  the  Sum  of  4-^54  and  2 ^250.  Ans.  22^/2.  ! 

14.  Find  the  Difference  between  5\/9x3  and  3-v/x2.  Ans.  12x-[/x. 

15.  Find  the  Sum  of  2^/161 2 and  54 a.  A?is.  l\/2a. 

16.  Find  the  Difference  between  3 -\/ 10  and  5 \/l9.  Ans.  2\/l0- 

17.  Find  the  Sum  of  5-\/98x  and  10-\/2x.  Ans.  45-\/2x  j 

18.  Find  the  Difference  between  3a\/ 5 and  a^/5.  Ans.  2a\/5.  1 

19.  Find  the  Sum  of  a \/y2  and  35  \/y2.  Ans.  [a-\-3b)y/y2 . 

20.  Find  the  Difference  between  5\/5  and  2 ay/5.  Ans.  (5  — 2a)\/5 


21.  Find  the  Sum  of  (a+ 1)2  and  -y/4rt+4. 

Ans.  3 (a-f-1)* 

. \ 

22.  Find  the  Difference  between  y 14-a:  and  3 (1  + ^)2- 

Ans.  2 (1+ x)1  f 

23.  Find  the  Sum  of  2 (a— a;)2  and  V9a — 9x). 

1 

Ans.  5 (a—x)1  ; 

, , i 

24.  Find  the  Difference  between  y 2+?/ and  4 (2/+ 2)3. 

Ans.  3 (24-3/)* 

25.  Find  the  Difference  between  4 (l  + a:2)2  and  4 v^x2  + l. 

4 1 

Ans.  4 (l-)-x2— a;2  + y). 


MULTIPLICATION  ANl>  DIVISION  OF  SURDS. 


157 


MULTIPLICATION  AND  DIVISION  OF  SURDS. 

(241.)  1.  The  Product,  or  Quotient,  of  two  Surds  of  the  same  root , 
is  obtained  by  prefixing  the  product,  or  quotient,  of  their  coefficients  as 
a coefficient  to  the  product,  or  quotient,  of  the  radical  factors, — the 
latter  being  affected  with  the  same  fractional  exponent  or  radical 
sign. 

2.  Surds  of  different  roots  may  he  reduced  to  equivalent  ones  of 
the  same  root,  (239),  and  then  multiplied,  or  divided,  as  above.  But 

3.  Any  two  roots  of  the  same  quantity  may  be  multiplied  into 
each  other,  by  adding  together  their  fractional  exponents ; or  divided, 
the  one  into  the  other,  by  subtracting  the  exponent  of  the  divisor  from 
that  of  the  dividend. 


example.  • 

To  find  the  Product  of2'y/l0x3'v/2. 

Since  it  is  immaterial  in  what  order  the  four  factors  are  taken,  we 
may  take  them  in  the  order, 

2x3Vl0x  a/2;  which  gives  the  Product  6 a/20,  (234),  =12  a/5,  (23?^ 

The  Quotient  of  2a/10-^-3a/2  is  -|a/ 5,  since  this  quotient  multi- 
plied by  the  divisor  produces  the  dividend. 

EXERCISES 


On  the  Multiplication  and  Division  of  Surds. 


1.  Find  the  Product  of  5 a/8  X 3 y/5. 

Ans.  30a/10. 

2.  Find  the  Quotient  of  6a/54-^3a/2. 

Ans.  6 a/3. 

3.  Find  the  Product  of  a/108  X 2 a/6. 

Ans.  36  a/2. 

4.  Find  the  Quotient  of  2a/96-^-  a/54. 

Ans.  2§. 

5.  Find  the  Product  of  3 ffoax  X 4:\/20a. 

Ans.  120a  a/x. 

6.  Find  the  Quotient  of4A/l2an-2A/6. 

Ans.  2a/2 a. 

7.  Find  the  Product  of  a/3  ax  X 3 a /ax. 

Ans.  3axff3 

8.  Find  the  Quotient  of  12x2  =3a/ 4. 

Ans.  2a: a/3. 

9.  Find  the  Product  of  %/\8  X 5a/4. 

Ans.  lO-yAh 

10.  Find  the  Quotient  of  43v/72^2yi8. 

Ans.  2z^j  4. 

158 


MULTIPLICATION  AND  DIVISION  OF  SURDS. 


11.  Find  the  Product  of  5 a X 3 \/ a. 

By  reducing  the  surds  to  the  same  root , we  obtain 
5yV  and  3%/ a?,  (239). 

These  are  to  be  multiplied  together  as  before.  Ans.  1 5\/as. 

But  the  given  roots  of  the  same  quantity  a.  may  also  be  multiplied 
into  each  other,  by  adding  together  their  fractional  exponents  ^ and  -J. 
Thus, 

5a'-tx3a:i  = l5a6  — l5^/a5,  as  before. 

Either  of  these  two  methods  may  be  applied  to  roots  of  the  same 
quantity.  The  first  only  is  applicable  to  different  roots  of  different 
quantities. 

12.  Find  the  Product  of  3 x X 2 x. 

13.  Find  the  Quotient  of  2q/34-5^/3. 

14.  Find  the  Product  of  4q/3  X 2^/2. 

15.  Find  the  Quotient  of  8-y/a  — 4 V a. 

16.  Find  the  Product  of  ffaxX3y/'ax. 

17.  Find  the  Quotient  of  4 x3  -^2,ff  x. 

18.  Find  the  Product  of  5 X^V  10. 

19.  Find  the  Quotient  of  a3  \. 


Ans.  6^^- 
Ans.*^  \/3. 
Ans.  8-^108 
Ans.  2 1\/a. 
Ans.  3Va3a'3. 

Ans.  2x\J%. 
Ans.  i-\/250. 
Ans.  a2\/^. 


20.  Find  the  Product  of  (3  + 2-y^)  X (2  — 5). 

In  cases  of  this  kind,  in  which  a Surd  is  connected  with  another 
quantity  by  the  sign  + or  — , the  Multiplication,  or  Division,  must  be 
performed  as  on  polynomials. 

3 + 2'v/o 

2—  5 

G + 4-/5 

— 3-y/5  — 10 

6+  5 — l0  = -v/5 — 4. 

E ach  term  of  the  multiplicand  is  multiplied  by  each  term  of  the 
multiplier,  and  the  partial  products  are  then  added  together. 

Observe  that  2 q/5  X — 5—  — 2 \/25  = — 10. 

21.  Find  the  Product  of  (2  + 3^2)  x(l+5-\/2). 


M«s.  13-/2  + 32. 


RATIONALIZATION  OF  SURD  DIVISORS. 


159 


22.  Find  the  Product  of  (4-  ff?>)  X (2  + 3^3). 

Ans.  1 0 -\/  3 — 1. 

23.  Find  the  Product  of  (5  + 2-y/6)  X (1  + 2 -y/ 6) . 

Ans.  12-\/6  + 29. 

24.  Find  the  Product  of  (1 — 4-^/7)  X (3  — 3 -\/l). 

Ans.  87-1 5 -v/7. 

25.  Find  the  Quotient  of  ( -\/ 20  + 1 2 ( -\/ o + -y/ 3) . 

Ans.  2. 


Rationalization  of  Surd  Divisors. 

(242  ) In  computing  an  approximate  value  of  an  irrational  nume- 
rical expression,  it  is  expedient  that  a surd  divisor  or  denominator  be 
made  rational. 

For  example,  suppose  we  wish  to  compute  an  approximate  value  of 

— — - , 2 divided  by  the  square  root  of  3. 

V 3 

If  we  extract  the  square  root  of  3,  for  a divisor,  a regard  to  accu- 
racy will  require  that  the  root  be  continued  to  several  figures,  and 
hence  will  arise  the  inconvenience  of  dividing  by  a large  number.  • 

' By  multiplying  both  terms  by  the  denominator,  we  have 

-7-  = — — — , in  which  the  divisor  is  rational. 

V3  -t/9  3 

The  value  will  therefore  be  found  by  taking  of  the  square  root  of 
12  ; and  by  this  method  the  computation  is  much  simplified. 

In  pursuance  of  the  object  at  present  in  view,  it  is  necessary 

To  iind  a Multiplier  of  a given  Surd  which  will  cause 
the  Product  to  be  Rational. 


(243.)  1.  A monomial  Surd  will  produce  a rational  quantity  by 

being  multiplied  into  itself  with  its  exponent  subtracted  from  a unit. 

A , X-J-  i 2. 

Thus  a3  multiplied  by  a 3,  or  a 3 Xa3  =a,  (241 . . .3). 

2.  A binomial  in  which  one  or  both  terms  contain  an  irrational 
square  root , will  produce  a rational  quantity  by  being  multiplied  into 
itself  with  a sisn  changed. 

Thus  (-/  3+-/  2)x(/3— / 2)  = 3 — 2=1. 

The  product  in  this  case  is  readily  found  on  the  principle,  that  the 
Product  of  the  sum  and  difference  of  two  quantities  is  equal  to  the  dif 
ference  of  the  squares  of  the  two  quantities. 


160 


RATIONALIZATION  OF  SURD  DIVISORS. 


3.  A trinomial  containing  irrational  square  roots  will  produce  a 
binomial  Surd  by  being  multiplied  into  itself  with  a sign  changed ; 
and  this  binomial  may  be  rationalized  as  above. 

These  principles  provide  for  the  most  useful  cases  of  the  subject 
under  consideration. — In  applying  them  to  the  rationalization  of  surd 
denominators,  both  terms  of  the  given  Fraction  must  be  multiplied  by 
the  same  quantity,  (81). 


EXERCISES 

On  the  Rationalization  of  Surd  Denominators 
2+-v/3 


1.  Reduce  ~ — to  a Fraction  having  a rational  denominator. 

. 2^9+3 

5 _ 3 

2.  Reduce  t0  a Fraction  having  a rational  denominator. 

15  — 5\/  5 


3.  Reduce 


4.  Reduce 


5.  Reduce 


2 a 


2-fl 


Ans. 

to  a Fraction  having  a rational  denominator. 


Ans. 


\a-\-2afl 


-3 


3x 
■fa  — V x 


to  a Fraction  having  a rational  denominator. 


Ans. 


3xf a-\-3xf  x 


a—x 


10 

t+  Vio 


to.  Reduce 


a-V 2 
4/2 


to  a Fraction  having  a rational  denominator. 

Ans.  iq^-io-y/io 
a2  — 10  ‘ 


to  a Fraction  having  a rational  denominator 


Ans. 


VS -2 


7.  Reduc-e- 


f3+  f2.-\- 1 


to  a Fraction  having  a rational  denominator 

Arts.  4+2^2— 2-^/6. 


INVOLUTION  AND  EVOLUTION  OF  SURDS. 


161 


INVOLUTION  AND  EVOLUTION  OF  SURDS. 

(244- ) The  Powers  and  Roots  of  irrational  quantities  are  obtained, 
or  indicated , according  to  the  general  principles  of  Involution  and 
Evolution  which  have  been  established  in  the  preceding  Chapter. 

We  present  here  however  a particular  case  of  the 

Square  Root  of  Binomial  Surds. 

(245.)  A Numerical  Binomial  of  the  form  a±-\/b  admits  of  a 
square  root  in  a rational  and  an  irrational  term,  or  two  irrational 
terms,  whenever  a2 — b is  a perfect  square. 

To  determine  the  method  to  he  pursued  in  this  case  of  evolution,  we 
must  find 

Formulas  for  the  Square  Root  of  <z±-/  b. 

The  square  of  the  sum  of  any  two  quantities,  is  equal  to  the  sum 
of  their  squares  + twice  their  product,  (59).  The  binomial  aA--f  b 
may  therefore  represent  the  square  of  the  sum  of  a rational  and  an  ir- 
rational numerical  term,  or  of  two  irrational  terms,  in  the  square  root; 
a representing  the  sum  of  the  squares  of  the  two  terms, — which  sum 
will  necessarily  he  rational, — and  y'  b representing  twice  the  product 
of  the  two  terms. 

In  like  manner  a — f b may  represent  the  square  of  the  difference 
of  a rational  and  an  irrational  numerical  term,  or  of  two  irrational  nu- 
merical terms,  in  the  square  root,  (60). 

If  therefore  we  take  x and  y to  represent  the  two  terms  of  the 
square  root  of  a±f  b,  we  shall  have, 

(1)  x2jry2=a] 

(2)  x +y  —V^+fb, 

(3)  x ~y  a — ytb  ; 

Multiplying  together  equations  (2)  and  (3),  we  have, 

(4)  x2 — y2  = -\/a2 — b.  (58.) 

Adding  together  the  (4)  and  (1),  and  also  subtracting  the  (4)  from 
.he  (1),  and  dividing  by  2,  we  shall  find, 

8* 


162 


INVOLUTION  AND  EVOLUTION  OF  SURD8. 


a+Va2—  b 


a — -y/  a 2 — b 


Extracting  the  square  root  of  each  of  these  equations, 

£ i \ a-^-  4/  a~  b 

v 2 

j \ a — V °2  — b 

and  y = \ . 

- 

Substituting  these  values  of  x and  y in  equations  (2)  and  (3), — 
and  interchanging  the  first  and  second  members, — we  have, 

(A)  V^Wl=  ; | 

(B)  _ y+v^IEI . 

These  are  the  Formulas  required.  The  right  hand  member  ol 
each  will  contain  at  most  but  two  irrational  terms,  when  [a? — b)  is  a 
perfect  square , that  is,  has  an  exact  square  root. 

EXAMPLE. 

To  find  the  Square  Root  of  6-\-2f  5,  or  6+q/20,  (236.) 

Substituting  6 for  a,  and  -\/20  for  -fb,  in  Formula  (A), 

V6+720=  y+t^° 

2 2 


(5) 

(6) 


And  since  -\/36  — 20  = -v/l6  = 4,  the  second  member  reduces  to 

\/^  + 


=1/5  + 1,  the  Root  required. 


The  root  -y/o+l  may  be  verified  by  squaring  it. 

Thus  ('/5+l)2=5+2q/5+l— 64-2-v/5. 
Formula  (B)  would  give  the  square  root  of  6 -2^5. 


INVOLUTION  AND  EVOLUTION  OF  SURDS. 


163 


EXERCISES 

On  the  Powers  and  Roots  of  Surds. 

It  will  be  readily  perceived  that  by  cancelling  the  exponent  or  sign 
of  a root,  that  root  is  raised  to  the  corresponding  power. 

Thus  the  square  of  -fa  is  a ; the  cube  of  q/a  is  a , &c. 

1.  Find  the  Square,  and  also  the  square  root,  of  5q/4. 

The  square  of  5 X 4 3 is  25  X43  =25  V"  16,  (200). 

"Reducing  this  result  to  its  simplest  form,  we  have 

25  V 1 6=25  ^8  ^2  = 50  V 2,  (237). 

JL  . 1 1 6 • 

The  square  root  of  5x43  is  5~  x46,  or  y'S  V 4,  (208). 
Multiplying  together  the  two  factors  of  this  root,  wre  have 

v/5V4  = Vl25V4  = V500,  (241  ...  2). 

2.  Find  the  Square,  and  also  the  square  root,  of  9 q/3. 

Ans.  81  and  3 q/3. 

3.  Find  the  Cube,  and  also  the  cube  root,  of  3q/2. 

Ans.  54-\/2,  and 

4.  Find  the  Square,  and  also  the  square  root,  of  2 q/3. 

Ans.  4 q/9>  and  ^/24. 

5.  Find  the  Cube,  and  also  the  cube  root,  of  3a  q/3. 

Ans.  81a3  q/3,  and  y/27a3. 

6.  Find  the  Square,  and  also  the  square  root,  of  x 2 q/ 5. 

Ans.  a;4  q/25,  and  x \/lT. 

7.  Find  the  Square,  and  also  the  square  root,  of  3 q/n+T. 

Ans.  9 (a  + 1),  and^/9  (a+l). 

8.  Find  the  Square,  and  also  the  square  root,  of  4 q/l  — x. 

Ans.  16  ^/(l — x)2,  and  2^1 — x 

9.  Find  the  Cube,  and  also  the  cube  root,  of  2 q/a2 — x. 

Ans.  8 q/(a2  — x)  3,  andj/4  (a2—  x). 

10.  Find  the  Square,  and  also  the  square  root,  of  3 y/2 — a. 

Ans.  9 12 -a).  and  ft  9' (2 -a). 

11* 


164 


EXERCISES. 


11.  Find  the  Square  of  2+/3,  and  of  2-/3. 

These  squares  may  he  found  by  multiplying  each  binomial  into 
itself,  or,  more  readily,  by  applying  the  propositions  relating  to  the 
squares  of  the  sum  and  the  difference  of  two  quantities. 

Ans.  7 + 4/3,  and  7-4/3. 

12.  Find  the  square  of  3 + 2/5,  and  of  5 — 3ff~- 

A)is.  12/5  + 29,  and  43 — 30/2^ 

13.  Find  the  Square  of  V" 5 + 2/ a,  and  of  V 3 — 5/ x. 

Ans.  5 + 2/a,  and  3— 5/z. 

14.  Find  the  Square  of  / 2+3/ 5,  and  of  / a — 2x x. 

Ans.  6/10  + 47,  and  a — 4^/ax+4z3. 

15.  Find  the  Cube  of  ^1+5/4,  and  of  Va  — 3/a. 

Ans.  1 + 5/4,  and  a — 3/a. 

16.  Find  the  Square  of  /3  + a/3,  and  of  /3  — 3/a- 

Ans.  3 + 6a  + 3a2,  and  3 — 6/3^_|_ga> 

17.  Find  the  Cube  of  2 + 2/a,  and  of  2 — a/2. 

Ans.  8 + 24a  + (24  + 8a)/a,  and  8+  12a  — (12a  + 2a3)/2. 

18.  Find  the  Cube  of  (/a_ and  of  (/l  — 

Ans.  /a+/ 2,  and  v 1 — /yT 

19.  Find  the  Square  roo£  of  a + 2 /ax +x,  (218). 

Ans.  /a+  / x. 

20.  Fijrd  the  Square  root  of  a + 2/a+l,  and  of  3 + 2a/3  + a2. 

Am.  /a  + 1,  and  /3  +a. 

21.  Find  the  Square  root  of  x — 2/x+l,  and  of  5 — 2x/ 5 + .r2. 

Ans.  -/x — 1,  and  /5—  x. 

22.  Find  the  Square  root  of  23  + 8/7,  ( 245  ).  Formula  A. 

Ans.  4+/7. 

23.  Find  the  Squaie  root  of  194  8/3,  and  of  7 — 2/l0- 

+as.  4+  /3,  and  /o—  v 2- 


IMAGINARY  QUANTITIES. 


165 


IMAGINARY  QUANTITIES. 

(246.)  An  even  root  of  a negative  quantity  is  impossible,  and  the 
symbol  of  such  a root  is  therefore  called  an  imaginary  or  impossible 
quantity,  in  contradistinction  to  real  quantities. 

Thus  yj  — 4,  the  square  root  of  — 4,  is  imaginary,  since  there  is  no 
quantity  whose  square  is  — 4 ; and  for  a like  reason  -y/^16  is  ima- 
ginary. 

An  Imaginary  Quantity  results  in  calculation  from  some  impossi- 
bility in  the  conditions  of  a Problem,  and  may  therefore  be  regarded  as 
a sign  of  such  impossibility. 

As  an  instance  of  this,  suppose  it  were  required  To  find  a number 
whose  square  subtracted  fir om  5 shall  leave  9. 

If  x represent  the  required  number,  the  Equation  of  the  problem 
will  be 

5 — a;2  =9. 

From  this  equation,  x2  = — 4 ; and  hence  x = yj  — 4 

The  value  of  x being  imaginary  or  impossible,  shows  that  the 
problem  is  impossible ; that  is,  that  there  is  no  number  whose  square 
subtracted  from  5 will  leave  9. 

Imaginary  Quantities  may  become  the  subjects  of  calculation  like 
real  quantities.  Thus  we  may  wish  to  verify  an  imaginary  value  of 
the  unknown  quantity  in  an  Equation,  by  subjecting  this  value  to  all 
the  operations  which  are  performed  on  its  symbol  in  the  equation. 

Calculus  of  Imaginary  Quantities. 

(247-)  All  the  principles  which  have  been  established  for  the  cal- 
culus of  radical  quantities  are  applicable  to  imaginary  quantities,  ex- 
cept that  the  ambiguous  sign  ± does  not  precede  the  product  of  two 
imaginary  quantities,  as  is  the  case,  in  general,  with  even  roots  of  real 
quantities. 

The  sign  which  affects  the  Product  of  two  imaginaries  may 
always  be  determined  by  means  of  imaginary  and  real  factors  into 
which  such  quantities  may  be  resolved. 


Ifa6 


IMAGINARY  QUANTITIES. 


(243.)  An  Imaginary  Quantity  may  always  be  resolved  into  the 
like  root  of  ( — 1)  multiplied  into  the  like  root  of  a positive  quantity 
which  is  equal  to  the  negative  quantity  in  the  given  imaginary. 

Thus  —4  being  equal  to— 1x4,  we  have  y/  — 4 = y/ — 1 Xy/4  1 

and  — a being  equal  to — 1 X a,  we  have  y / — a = y/  — 1 x y/a.(234-) 

By  means  of  this  transformation  it  may  be  shown,  that 

(249  ) The  Product,  of  two  imaginary  squire  roots  is  the  negative 
square  root  of  the  product  of  the  two  quantities  under  the  -y/,  if  the 
given  roots  are  preceded  by  like  signs,  -f-  or  — ; otherwise,  it  is  the 
vositive  square  root  of  that  product. 

For  example,  -y/  —a  . y / —b=  — yfab; 

and,  -y / —a  . — -y/  — b—  + i/ab. 

On  general  principles  of  calculation,  the  Product  of  the  square  roots 
of  —a  and  —b,  is  equal  to  the  square  root  of  the  product  ab . which 
would  be  -\-yf ab,  or  else  — -y f ab,  (215). 

But  the  ambiguity  in  the  sign  to  be  prefixed  to  an  even  root  in 
general,  is  removed  when  we  know  the  factors  which  entered  into  the 
composition  of  the  quantity  whose  root  is  considered.  When  a2,  for 
example,  is  known  to  have  been  derived  from  ax  a,  the  square  root  of 
a2  is  a,  and  not  — a. 

V 

To  determine  the  sign  of  -y fab  in  the  first  example,  we  have 
\/  — a . y/  — b — yj  — 1 . y /a  . y/  — 1 . y / b,  (24S) 

= y/  — 1 ■ y / — 1 . y/ ay/ b 

— {V  — l)2y^ab=  — ly/~ab,  or  — y fab. 

In  the  second  example,  we  have 

yf—a  . — y/^Q  = y/^l  . y/a  . — y/^1  . y /b,  (248), 

= y/  — "l  . — y/  — 1 . y/ ay/ b 
— — (y/  — l)2y fab—  — ( — l)y rabzii-fyfab. 

In  this  example  the  square  (y/  — l)2  has  the  sigu  — before  it,  be 
cause  it  results  from  multiplying  y ' — 1 by  — \ — 1,  (42);  and 
— ( — 1)  becomes  +1  by  changing  the  sign  in  subtracting. 


167 


CHAPTER  X. 

QUADRATIC  AND  OTHER  EQUATIONS. 

(250.)  A Quadratic  Equation,  or  an  equation  of  the  second  de- 
cree, is  one  in  which  the  highest  power  of  the  unknown  quantity  is  its 
square,  or  second  power,  as 

3z2  = 12;  or  3x2+4a:=20. 

A Cubic  Equation,  or  an  equation  of  the  third  degree,  is  one  in 
which  the  highest  power  of  the  unknown  quantity  is  its  cube,  or  third 
power  ; and  in  like  manner  is  defined  a Biquadratic  Equation,  or  an 
equation  of  the  fourth  degree,  and  so  on. 

An  Equation  containing  two  or  more  unknown  quantities  is  of  the 
degree  which  corresponds  to  the  greatest  number  of  unknown  factors 
in  any  of  its  terms. 

Thus  oxy-\-y=a  is  an  Equation  of  the  second  degree,  its  first  term 
containing  the  unknown  factors  xy. 

And  x2y — y=b  is  an  equation  of  the  third  degree.,  its  first  term 
containing  three  unknown  factors  xxy. 

Pure  and  Affected  Equations. 

(251 .)  A Pure  Equation  is  an  equation  which  contains  hut  one 
power  of  the  unknown  quantity ; and  is  a Simple  Equation,  a Pure 
Quadratic,  or  a Pure  Cubic,  &c.,  according  to  its  degree. 

. 

Thus  3a;2  is  a pure  quadratic  ; a:3  = 64  is  a pure  cubic. 

An  Affected  Equation  is  one  which  contains  different  powers  of 
the  unknown  quantity.  When  these  powers  are  in  regular  ascension, 
beginning  with  the  first  power,  the  Equation  is  also  called  a complete 
equation. 

Thus  a;2-f  3x  = 10  is  an  o.ffected,  and  also  a complete  quadratic  , 
x3—  2.r2=7-5,  or  x3 — 3.r=110,  is  an  affected  cubic;  and  x3  -\-oX2 
| 4.t;  = 28  is  a complete  cubic. 


168 


GENERAL  PROPERTIES  OF  EQUATIONS. 


Roots  of  Equations. 

(252.)  A Root  of  an  Equation  is  a value  of  the  unknown  quantity 
in  the  equation.  It  will  presently  be  shown  that  an  equation  of  the 
2d  degree  has  two  roots,  of  the  3d  degree  three  roots,  and  so  on. 

In  the  Simple  Equation  3*=  15,  the  value  of  the  unknown  quantity 
x is  5 ; then  5 is  the  root  of  the  equation. 

It  is  evident  that  in  a Simple  Equation  there  can  be  but  one  valu« 
of  the  unknown  quantity  that  will  satisfy  the  equation.  An  equation 
of  the  1st  degree  has  therefore  hut  one  root. 


General  Properties  of  Equations. 

1.  Divisors  of  an  Equation. 

(253.)  If  a he  a root  of  an  Equation  of  any  degree,  containing  hut 
one  unknown  quantity,  x ; the  equation — with  all  its  terms  transposed 
to  one  side — will  be  divisible  by  a: — a. 

Let  a be  a root  of  the  Equation 

x2  -\-mx=s  ] 

then  will  the  equation  x2+mz — s = 0 be  divisible  by  x -a 

For  let  r be  the  remainder , if  any,  after  the  quotient  q has  bee* 
obtained ; then  will 

x2 -\-mx — s = (x  — a)q-j-r  — 0 ; 

the  dividend  being  equal  to  the  remainder  added  to  the  product  of  the 
divisor  and  quotient. 

But  a being  a value  of  x,  (252),  we  have  x— a = 0 ; then  the  {x—a) 
X q is  0,  (43) ; and  consequently  r = 0 ; that  is,  the  division  will  leave 
no  remainder. 

The  preceding  demonstration  is  applicable  to  an  equation  of  the 
third,  fourth,  or  any  higher  degree. 

(254.)  Conversely,  If  an  Equation  of  any  degree,  containing  but 
one  unknown  quantity,  x, — with  all  its  terms  transposed  to  one  side — 
be  divisible  by  x—a ; then  a will  be  a root  of  the  equation. 

This  is  evident  from  considering  that  the  divisibility  of  the  Equation, 
as  shown  above,  depends  on  the  condition  that  x—a  — 0,  or  that  a is  a 
value  of  r. 


GENERAL  PROPERTIES  OF  EQUATIONS. 


169 


2.  Number  of  Roots  of  an  Equation. 

(255.)  Every  Equation  containing  but  one  unknown  quantity,  has 
just  as  many  roots  as  there  are  units  in  the  exponent  of  the  highest 
power  of  the  unknown  quantity  in  the  equation. 

Let  a represent  a root  of  the  cubic  Equation 
x3  -\-mx2  -\-nx-s. 

Transposing  s to  the  first  side  of  the  equation,  we  have 
x3+mx 2 +nx — s=0. 

Dividing  this  equation  hy  x—a,  (253),  we  shall  obtain  an  equation 
of  the  second  degree,  which  may  be  represented  by 
x2  -\-px — q=  0. 

Let  b be  a root  of  this  equation.  Dividing  the  equation  by  x-  b 
we  shall  obtain  an  equation  of  the  first  degree,  represented  by 

x — U — 0. 

The  binomials  x — a,  x — b , and  x — u,  which  represent  the  two  di 
visors  and  the  last  quotient,  are  the  factors  of  the  dividend. 

x3  +ffli2  -\-nx — s. 

The  original  equation  is  thus  resolved,  representatively,  into 
(x — a)(x — b){x — u)  =r0. 

Since  this  equation  is  divisible  by  each  of  these  three  binomial  fac 
tors,  it  follows  that  a,  b,  and  u are  three  roots  of  the  equation,  (254). 

And  since  the  given  equation — being  of  the  3d  degree — cannot  he 
resolved  into  more  than  three  binomial  factors,  each  containing  the 
first  power  of  x,  it  cannot  have  more  than  three  roots. 

The  same  method  of  demonstration  will  show  the  proposition  to  be 
true  for  an  Equation  of  any  other  degree. 

The  several  roots  of  an  Equation  are  not  necessarily  unequal, 
though  such  will  usually  be  found  to  be  the  case.  The  preceding  Pro- 
position shows  that  an  Equation  may  be  resolved  into  binomial  fac.toxs, 
each  containing  a root  of  the  equation.  Two  or  more  of  these  roots 
may  be  equal  to  each  other. 

In  the  Equation  x2  — 5x  + 6 = (x — 2)(x  — 3)  = 0, 
the  two  roots  are  2 and  3. 

In  the  Equation  x3 — 7a;2  + l6a: — I2  = (a:  — 2)(x  — 2)(x — 3)  = 0 
the  three  roots  are  2,  2,  and  3. 


170  . 


SOLUTION  OF  PURE  EQUATIONS. 


Solution  of  Pure  Equations  of  the  Second  and 
Higher  Degrees. 

The  Equations  belonging  to  this  class  are  those  which,  in  their 
simplest  forms,  contain  but  one  power  of  the  unknown  quantity 

RULE  XXII. 

(256.)  Ear  the  Solution  of  a Pure  Equation. 

1.  Reduce  the  Equation  to  the  form 

xn=s  ; in  which  xn  must  he  positive. 

2.  Extract  that  root  of  both  sides  of  the  equation  which  corresponds 
to  the  power  of  the  unknown  quantity 

e x A M p l e . 

To  find  the  value  of  x in  the  Equation 
x2 

t-5  = 2x2  — 23. 

4 

Clearing  the  equation  of  its  Fraction,  we  have 
z2  + 20  = 8a:2—92. 

Transposing,  and  adding  similar  terms,  we  find 
—7x2  = — \12. 

Dividing  both  sides  of  this  equation  by  the  coefficient  — 7, 
x2  = 16. 

Extracting  the  square  root  of  both  sides, 

x=  ±4,  (215). 

The  two  values  of  x are  thus  found  to  be  4 and  — 4,  (255),  either 
of  which  will  satisfy  the  given  Equation. 

It  is  evident- that,  in  a Pure  Equation  of  the  second  degree,  the 
two  values  of  the  unknown  quantity  will  always  be  equal , with  con 
tranj  signs. 


The  unknown  quantity  may  enter  an  Equation  in  a surd  expres- 
sion, which  it  will  be  necessary  to  rationalize  in  the  solution  of  the 
equation. 

Thus  in  the  equation  -fx-j-  y/ 1 -\-x=a,  it  would  be  necessary  to 
rationalize  x,  that  is,  to  clear  it  of  the  radical  sign  before  the  value 
of  x could  be  determined. 

The  following  observations  will  assist  the  student  in  the 


RATIONALIZATION  OF  SURD  QUANTITIES  IN  AN  EQUATION.  J 


Rationalization  of  Surd  Quantities  in  an  Equation. 

(257.)  A Surd  quantity  in  an  Equation  will  be  rationalized  by 
transposing  all  the  other  terms  to  the  other  side  of  the  equation,  and 
raising  both  sides  to  the  power  corresponding  to  the  indicated  root. 

To  rationalize  x in  the  Equation 

Vx+1  —a—b. 

By  transposition,  V ze+  1 =<z+5 
Squaring  both  sides,  x-\- 1 =a2 -\-2ab-\-b2 . 

(258.)  Two  Surds  in  an  Equation  may  be  rationalized  by  successive 
involutions, — in  the  first  of  which  it  will  generally  be  expedient  to 
make  one  of  the  Surds  stand  alone  on  one  side  of  the  equation. 

To  rationalize  x in  the  Equation 

V %+  V 1 +x=a. 


By  transposition, 
Squaring  both  sides, 

By  transposition, 
Squaring  both  sides, 


V x=a—V  1 + a;. 


x=za 2 — 2a  V l + a;+  \-\-x. 


2 aV  l + a=a2  + l. 

4a2(l  + a)  = «4  + 2a2  + l. 


(259.)  When  an  Equation  contains  a Fraction  whose  terms  are 
both  irrational,  it  will  sometimes  be  expedient  to  rationalize  its  de- 
nominator before  clearing  the  equation  of  the  fraction. 

To  rationalize  x in  the  Equation 
Wx 

— - — — —a. 

l+yA 

Multiplying  both  terms  of  the  Fraction  by  1 — -y/ x,  (243. ..2), 

■y/  X — X 


1 — X 


-a. 


Clearing  this  equation  of  its  Fraction,  and  transposing,  we  have 
y tx=za—ax-\-x. 

The  Surd  in  this  equation  will  be  rationalized  by  squaring  both 
sides,  as  in  the  preceding  examples. 


172 


EXERCISES. 


By  the  preeedikg  methods  we  may  generally  rationalize  one  or 
more  Surds  containing  the  unknown  quantity  in  an  Equation.  Other 
expedients,  however,  such  as  the  extraction  of  roots  in  possible  cases, 
in  the  course  of-ihe  operation,  will  sometimes  be  requisite ; hut  tliesp 
must  be  left  to  the  care  and  skill  of  the  computer. 


EXERCISES 

On  Rationalization , and  Pure  Equations. 

1.  Find  the  value  of  x in  the  equation 

24  — -[/2x2  +9  = 15.  Ans.  X — dbb 

2.  Find  the  value  of  x in  the  equation 

13  — -v/3x2  + 16  = 5.  Ans.  *=±4 

3.  Find  the  value  of  x in  the  equation 

35+ V/z-5=40.  Ans.  a:=130 

4.  Find  the  value  of  x in  the  equation 

l + 2'\/x=v/4z+21.  Ans.  x=25 

5.  Find  the  value  of  x in  the  equation 

x—2>2  — ^ x—\y/2>2-  Ans.  z=50 

6.  Find  the  value  of  x in  the  equation 

3 + l/^+lX  V x — 4 = 10.  Ans.  x=±\/65. 

7.  Find  the  value  of  x in  the  equation 


a-{--\/x — 3 X V a;+3=4«. 


10 


8.  Find  the  value  of  x in  the  equation 

V^+v^+5‘=pj=- 

9.  Find  the  value  of  x in  the  equation 

4-v/6.r — 9 \/6x— 2 

4-\/6.r+6  -v/6x+2 

*10.  Find  the  value  of  x in  the  equation 
Vx+c  b V 


Ans.  x=±2>-fa2  + 1. 


\Zx+c—Vx — c 


= 3. 


Ans.  x—\ | 


Ans. 

cb2 +c 


Ans.  x— 


2b  ' 


SOLUTION  OF  COMPLETE  EQUATIONS. 


173 


Solution  of  Complete  Equations  of  the 
Second  Degree. 

These  Equations  in  their  simplest  forms  contain  no  other  power  of 
the  unknown  quantity  than  its  square  and  first  power. 

The  value  of  the  unknown  quantity  will  he  found  by  the  following 

RULE  XXIII. 

(260.)  For  the  Solution  of  a Complete  Equation  of  the  Second 
Degree. 


1.  Reduce  the  Equation  to  the  form 

x2  -\-bx=s  ; in  which  x2  must  he  positive. 

2.  Awd  the  square  of  half  the  coefficient  of  x,  in  the  second  term, 
to  each  Mde  of  the  equation : — the  first  side  will  then  be  a perfect 
square. 

3.  Ei  wact  the  square  root  of  each  side,  and  the  result  will  he  a 
Simple  Equation, — from  which  the  value  of  x may  readily  he  found. 

example. 


To  find  the  value  of  x in  the  equation 

2>x  — 5 5x- 

2x-\ — =5x- 


2 i x~3 

Clearing  the  equation  of  its  fractions,  transposing,  and  adding  sim- 
ilar terms,  we  shall  find 

— 3x2  f-14tX= — 5. 

Dividing  both  sides  of  this  equation  by  — 3, 

„ 14*  5 

X ~ ~ =3' 

Adding  the  square  of  half  the  coefficient  of  x,  in  the  second  term, 
to  each  side, — which  is  called  completing  the  square , 

2 14a;  49  _ 5 49  _ 64 

X 3~  + '9'-3  + T — ~9' 

Extracting  the  square  root  of  each  side, 

7 8 

x — — = ± - . 

3 3 

m 7 8 7 8 1 

Whence  x=  — — =5  ; or  *= = . 

3 3 ’ 3 3 3 


(255). 


174  SOLUTION  OF  COMPLETE  EQUATIONS. 

Either  of  these  two  values  of  x,  5 or  — will  satisfy  the  given 
equation ; and  each  of  the  binomials  x — 5 and  2+-^  will  divide  the 
equation 

*2“^r-§=°-  (253)-  ! 

By  performing  the  division  it  will  he  found  that  the  left  hand  side 
of  this  equation,  is  the  product  of  the  two  binomials  ; that  is, 

(x  5)(x-\-jsr)=x2  ~ ~ =0. 

On  the  preceding  Rule  we  remark, 

1.  The  method  of  completing  the  square  in  the  first  member,  re-  ji 
suits  from  the  composition  of  the  square  of  a Binomial. 

Thus  the  square  of  the  binomial  a-\-b  is  a2  -\-2ba-\-b2 , in  which  b t 
is  half  the  coefficient  of  a in  the  second  term. 

2.  If  in  reducing  the  equation  to  the  required  form,  the  first  term  1 1 
should  become  — x2,  the  signs  of  all  the  terms  in  the  equation  must  be:  I 
changed  (117),  before  completing  the  square, — otherwise  the  root  of  1 
the  first  side  would  be  imaginary , (246). 

3.  The  square  root  of  the  first  side  of  the  equation — after  the  eom-1 
pletion  of  the  square — will  always  be  the  square  root  of  the  first  termfi I 

f-  or  — half  the  coefficient  of  x in  the  second  term,  according  as  the  I 
second  term  is  + or  — 

Hence,  without  completing  the  square  on  the  first  side,  we  may  1 
shorten  the  solution,  by  observing  that 

4.  In  an  equation  of  the  form  x2  -f-6.r=s,  the  value  of  x is  half  the 
co-efficient  of  2 in  the  second  term,  taken  with  a contrary  sign,  + the 
square  root  of  (the  second  member  of  the  equation  + the  square  of  said  > 
half  co-efficient). 


14#  5 

Thus  from  the  equation  x2 T— — — , we  immediately  find 


*=Z+x/E^_l+x/6i=Z+? 

3 V 3 ' 9 3 — -V  9 3-3 


EXERCISES. 


175 


EXERCISES 


On  Quadratic  Equations  with  One  Unknown 
Quantity. 


1. 

2. 

3' 

4. 

5- 

6 

, 


9. 

10. 

11. 

12. 

13. 

14. 
15 


Find  the  value  of  x in  the  equation 

x 2 — 15  = 45 — 4a;. 

Find  the  value  of  x in  the  equation 

22  + 10  = 65  + 62. 

Find  the  value  of  x in  the  equation 

222  + 82— 30  = 60. 

Find  the  value  of  x in  the  equation 

3a:2—  32+9=8^. 

Find  the  value  of  in  the  equation 

5a:2 + 4x  — 90  = 114. 

Find  the  value  of  x in  the  equation 

+c2  — ^2+2  = 9. 

Find  the  value  of  x in  the  equation 

222  + 122+36  = 356.  Ans.  2=10  or — 16 

F ind  the  value  of  x in  the  equation 

. .„  36—  x 

4a: — 46=  . 


Ans.  2=6  or  — 10. 
Ans.  2 = 11  or  — 5. 
Ans.  2=5  or  —9. 
Ans.  X—\  or  I-. 


Ans.  2 = 6 or  — 6|. 


Ans.  a:  = 4 or  — 3+ 


Find  the  value  of  x in  the  equation 

8tc2 + 6 = 72+ 171. 
Find  the  value  of  x in  the  equation 

5,_23  = 2-^. 

x 

Find  the  value  of  x in  the  equation 

322  + 6 = 32+5^-. 
Find  the  value  of  x in  the  equation 
22  x 

T“  3 +2°*=42* 
Find  the  value  of  x in  the  equation 

8 — x x — 2 ( 2^—11 

~2~  = ~6  + 2 — 3 

Find  the  value  of  x in  the  equation 

lx  — 8 

2+4  = 13 . 

x 

Find  the  value  of  x in  the  equation 

2 — 3 a 9 (b — a) 


Ans.  2=12  or  — f-. 


Ans.  2=5  or  — 4^. 
Ans.  2=5  or  — 1. 
Ans.  2=-^*  or  J. 
Ans.  2=7  or  — 6^. 

Ans.  2=6  or 

Ans.  2=4  or  — 2 


2 


Ans.  2=3 b,  or  3 (a—b) 


176 


EXERCISES. 


Another  Method  of  Solving  Quadratic  Equations. 

(261.)  A Binomial  of  the  form  ax2  -\-bx  will  he  made  a perfect 
square  by  multiplying  it  by  4 times  the  coefficient  of  x2 , and  adding 
the  square  of  the  given  coefficient  of  x. 

Thus  {ax2  -\-bx)\a-\- b2  =4a2x2  -\-\abx-\-b2  — {2ax-\- b)2,  (59). 

To  apply  this  principle  to  solution  of  the  Equation 
3 a:2  — 2x=G5. 

Multiplying  both  sides  of  the  equation  by  4 times  the  coefficient  3, 
a nd  adding  the  square  of  the  coefficient  2 to  both  sides,  we  have 
36a:2  — 24*  + 4 = 780 + 4 = 784. 

Extracting  the  square  root  of  both  sides,  we  find 

i 03  | 2 

6a:  — 2=  ±28;  which  gives  x=z~  — — — 5 or  — 4-^. 

The  square  root  of  the  first  side  of  the  Equation  prepared  as  above, 
will  be  x multiplied  into  twice  the  given  coefficient  of  a:2,  + or  — 
the  given  coefficient  of  x in  the  second  term,  according  as  this  term  is 
-f-  or  — . This  is  evident  from  the  preceding  illustration,  (261.) 

From  these  principles  we  derive 

RULE  XXIV. 

(262.)  To  Reduce  an  Equation  of  the  form  ax2+bx=s  to  a 
Simple  Equation. 

1.  Double  the  coefficient  of  x2,  and  divide  the  first  member  by  x. 

2.  Multiply  the  second  member  by  4 times  the  given  coefficient  oi 
x 2,  add  the  square  of  the  given  coefficient  of  x,  and  extract  the  square 
root  of  the  sum. 

Applying  this  Rule  to  the  numerical  equation 
3a:2  — ox-50, 

we  immediately  obtain  6a;  — 5 = ± -v/50  X 12  + 2o=  ± -y/625.  : 

When  the  coefficient  of  x2  is  unity,  the  multiplier  of  the  second 
member  will  be  simply  4 ; and  when  the  coefficient  of  x is  unity,  the 
square  to  be  added  after  multiplying  the  second  member  is  1. 

This  method  of  solving  a Qmadratic  is  preferable  to  the  one  first 
given,  whenever  the  coefficient  of  x would  give  rise  to  a fraction  in 
dividing  the  Equation  by  the  coefficient  of  x2,  or  in  completing  the 
square,  according  to  that  method. 


EXERCISES. 

16.  Find  the  value  of  x in  the  equation 

2>x2-\-2x  — 9 = 76. 

17.  Fin'd  the  value  of  x in  the  equation 

2a:2  — 14a:+2  = l8. 

18.  Find  the  value  of  x in  the  equation 

x 2 — 122+50  = 0. 

19.  Find  the  value  of  x in  the  equation 

\x2  — 12^-=4-a;+ 15^-. 

20.  Find  the  value  of  x in  the  equation 

3a:2—  f=5i+2a;. 

21  Find  the  value  of  x in  the  equation 

5x 
6 

22.  Find  the  value  of  x in  the  equation 

. t+1=13-— . 

x 

23.  Find  the  value  of  x in  the  equation 


*2-  — +1=0. 


177 

Ans.  2=5,  or  — 5|-. 
Ans.  x=8,  or  — 1. 
Ans.  x=6dz\/ — 14, 
Ans.  2=8,  or  —7. 
Ans.  a;=|±^yi9. 

Ans.  x=%,  or  + 
Ans.  2=4,  or  —2. 


ju 

y + — — 34|=0.  Ans.  2=9,  or  —Ilf. 


24.  Find  the  value  of  x in  the  equation 


n X 2 X , , 

cc2  Ans.  a^-jLijL/97. 


Ans.  2=3,  or  1{2. 
Ans.  x—6±^ — 4. 


2 3 3 

25.  Find  the  value  of  x in  the  equation 

10  14  — 2a; 

— o± 

x x 2 9‘ 

26.  Find  the  value  of  x in  the  equation 

x2 

— — + 32+3  = 13. 

27.  Find  the  value  of  x in  the  equation 

„ 17.2  \lx 

x2-j = 4.  Ans.  X—  — 4,  or  — 8 

4 4 2 

28.  Find  the  value  of  x in  the  equation, 

3a: — 3 3a:  — 6 

ox — =2*-) — . Ans.  x =4,  or  -1. 

a:  — 3 2 

29.  Find  the  value  of  2 in  the  equation 

2+4  7 — x 42+7 

~ 3 ^c~3  =~ 1 ^ — 9 — ‘ ^ nS ‘ x~21’  or  5 

30.  Find  the  value  of  a:  in  the  equation 

x+a  32 


x — a 


= 0. 


Ans.  x=d=ay' — + 


178 


EXERCISES. 


Equations  in  which  the  Unknown  Quautity  is  contained 
in  a Surd  Expression. 

31.  Find  the  value  of  x in  the  equation 

5 + V'a;3  + 36  = 15- 

By  transposition,  y'a3+36  = 15  — 5 = 10. 

Squaring  both  sides,  a;3  + 36  = 100;  (257). 

from  which,  a:3  =64. 

We  have  now  a pure  cubic  equation,  in  which  x has  necessarily 
three  values  or  roots,  (255). 

Extracting  the  cube  root  of  each  side  of  the  last  equation,  we  find 
;r  = 4,  which  is  one  value  of  x. 

To  find  the  other  two  values  of  x , we  must  reduce  the  cubic  equa- 
tion to  a quadratic  by  division,  (253). 

Dividing  each  side  of  the  equation  x3 — 64  = 0 by  a:  — 4,  we  find 

x2  +4a:+ 16  = 0,  or  x2-j-4x=  — 16  ; hence  x— — 2±y/  — 12. 

We  have  thus  found  3=4,  or  — 2-b-y/  — f2,  or  — 2 — \/^12  ; the 
first  value  being  real,  the  other  two  imaginary . 

Each  of  these  imaginary  values  of  x,  as  well  as  the  real  value  4, 
will  satisfy  the  equation  a;3  =64. 

Thus  x being  equal  to  —2+  V — 12,  we  have,  (247),  (249), 
x- 2 = ( - 2 + \2 ) 2 = 4 - 4 /Hl2  - 1 2 = - S - 4 /^12 . 

and  a:3  =(  — 8 — 4-y/  — 12)(  — 2-fi  -y/  — 12)  = 16  + 48  = 64, 

In  like  manner  the  other  imaginary  value  of  x may  he  verified. 

32.  Find  the  value  of  x in  the  equation 

6 + V3*+4  = ll.  Ans.  x=7. 

33.  Find  the  value  of  x in  the  equation 

24—  V2x2  + 9 = \5.  Ans.  x=  ±6. 

34.  Find  the  value  of  x in  the  equation 

20  — ■V/a;3  + 40=4 

Ans.  .r=6  or—  3+V —27,  or  —3  — 7-27. 


EXERCISES. 


179 


35.  Find  the  value  of  x in  the  equation 

x+  y/X  = 20. 

From  this  equation  Ave  shall  find  a;  = 25  or  16.  The  value  25  will 
not  satisfy  the  original  equation  if  the  square  root  of  x be  restricted  to 
its  positive  value  ; but  this  root  is  ± y'.c,  and  25  satisfies  the  equation 
for  the  negative  root. 

Two  values  of  a:  may  be  found  in  each  of  the  next  three  Exercises  ; 
hut  only  that  value  is  given  in  the  Ans.  Avhich  corresponds  to  positive 
roots  in  the  given  Equation. 


36.  Find  the  value  of  x in  the  equation 

2Vx+  VxAA  = l3. 

37.  Find  the  value  of  x in  the  equation 

2Vx-{-V2x+l  = 


■\Z{2x-\- 1) 


38.  Find  the  value  of  x in  the  equation 


4 V 2 + 16=7  V^ar+lfi — x — 6. 
39.  Find  the  value  of  x in  the  equation 


Ans.  2=16. 


Am.  2=4. 


Am.  a=9 


=V4  q-  V 2x3  -f-22.  Ans.  2=12,  or  4. 


40.  Find  the  value  of  x in  the  equation 

2+y'(22— 9) 


(2-2)2  = 

By  rationalizing  the  denominator,  (243... 2),  we  shall  find 
(2-2)2  = (- 


x~  V(x2~ 9) 


.2_(^+ v(*2-9))2 


Extracting  the  square  root  of  each  side  of  this  equation,  we  have 


2-2  = 


x+  qA*2— 9) 


Ans.  2=5,  or  3 


41.  Find  the  value  of  x in  the  equation 

22  — 6z+9=  Ans.  2 = 5.  or  4 

x — -\/{x2  — 16) 

42.  Find  the  value  of  x in  the  equation 

^/23  + 37x(23  + 37p  = 64. 

Ans.  2 = 3,  or— f±^A/ — 3. 


180 


EXERCISES. 


Equations  of  a Quadratic  Form  with  reference  to  a Poioer 
or  Root  of  the  Unknown  Quantity. 

(263.)  Any  Equation  containing  the  unknown  quantity  * in  but 
two  terms — with  its  exponent  in  one  double  its  exponent  in  the  other 
— is  a Quadratic  with  reference  to  the  lower  power  of  x ; and  the 
value  of  such  power  may  be  found  accordingly. 

To  find  the  value  of  * in  the  equation 
*72+4*^=21. 

1 . i 

The  higher  fractional  power  x'1  is  the  square  of  the  lower  a:4  ; and 
the  equation  is  therefore  quadratic  with  reference  to  a:4. 

Completing  the  square,  we  have 

*^+4*4+4=21  +4  = 25. 

Extracting  the  square  root  of  each  side, 

*^+2=  ±5  ; 
o. 

from  which'*4  = 3,  or  — 7. 

By  raising  each  of  these  values  to  the  4th  power,  we  find 
*=81,  or  2401. 

The  first  of  these  two  values  of  x is  easily  verified.  In  verifying 
the  value  2401  it  must  be  observed  that  its  4th  root  is  — 7,  and  that 

4*4  is  therefore  — 28. 

When  * is  in  a fractional  power,  in  the  following  Exercises,  only  f 
that  value  will  be  given  in  the  An's,  which  satisfies  the  given  form  of 
the  Equation. — Imaginary  values  of  x are  also  omitted. 

Ans.  *=±4. 
Ans.  *=^/3.  j 
Ans.  *=4^/6. 

Ans.  *=3-l-j-. 

Ans  *=S 


43.  Find  the  value  of  * in  the  equation 

x4—  2*2  + 6 = 230. 

44.  Find  the  value  of  x in  the  equation 

z6  + 20*3  — 10  = 59. 

45.  Find  the  value  of  * in  the  equation 

2*4 -*2  + 20=23. 

46.  Find  the  value  of  * in  the  equation 

3aQ— 5*4  = — 1+ 

47.  Find  the  value  of  x in  the  equation 

6x3— 5*3+1184  = 0. 


EXERCISES 


181 


48.  Find  the  value  of  x in  the  equation 

(a;-j-12)^+(a:+12)4  = 6. 

In  this  equation  we  must  regard  the  binomial  (£+12)  as  the  un- 
known quantity  ; and  to  simplify  the  operation  we  may  represent  this 
binomial  by  y. 

Then  ?/'2+?/4  = 6 
which  gives?/4  = 2 or  — 3. 


By  restoring  the  binomial  value  of  y,  we  have 
(a:+12)4  = 2,  or  —3. 


Ans.  x=4. 


49.  Find  the  value  of  x in  the  equation 

(2a;+6)^— 6 = — (2*+6)^. 

50.  Find  the  value  of  x in  the  equation 

x2+  11  + •v/a:2  + ll=42. 

51.  Find  the  value  of  x in  the  equation 


(2x — 4): 


=1  + 


(2x  — 4)4  ' 


Ans.  x—5. 
Ans:  x=  ±5 
Ans.  x=3,  or  1. 


52.  Find  the  value  of  x in  the  equation 


4a:4  - 


-4a:3  + - =33. 


This  equation  may  be  reduced  to  the  form  of  a quadratic , thus. — 
The  first  two  terms  of  the  square  root  of  the  first  side  will  be  found  to 
be 

2a;2 — x ; and  the  remainder  will  be  — x2  ^ , which  may  be 
put  under  the  form  — i-(2x2  — x). 

Now  the  square  of  the  root  found,  and  the  remainder , are  together 
equivalent  to  the  first  side  of  the  equation  ; hence  we  have 

(2x2  —x)2  —\(2x2  —x)=33.  Ans.  x=2,  or— 1£, 


A Biquadratic  equation  may  be  reduced  to  the  form  of  a Quad- 
ratic, as  above,  whenever  the  remainder — after  having  found  the  first 
two  terms  of  the  square  root  of  the  first  side — can  be  resolved  into  two 
factors,  one  of  which  is  the  same  as  the  part  of  the  root  thus  found 


182  EXERCISES. 

53.  Find  the  value  of  z in  the  equation 

x3 — 8a2  + 19a;  = 12. 

Transposing  the  12,  and  multiplying  by  x,  we  have 
x4  — 8a3  + 19a;2  — 12a  = 0. 

We  have  now  a Biquadratic  equation  which  may  be  reduced  to 
the  form  of  a Quadratic  by  the  method  just  exemplified. 


We  fi.ad  the  first  two  terms  of  the  square  root  of  the  first  side  of 
the  Equation,  by  the  common  Rule. 

x4  — 8a;3  + 19a2  — 12a:  (a2  — 4a; 

/y.4 


2a;2  — 4a;  /-8a3  + 19a2 
/ — 8a;3  + 16x2 


3a;2  — 12a:. 

The  remainder,  3a:2  — 12a:,  may  be  resolved  into  3 (a:2 — Ax),  in 
which  the  binomial  factor  is  the  same  as  the  part  of  the  square  root 
above  found. 

Then  (a:2  — 4a)2  +3  (a:2  — 4a)  = 0. 

Ans.  a=4,  3,  or  1. 

54.  Find  the  value  of  x in  the  equation 

x4  — 2a3+a  = 51l2. 

Ans.  a = 9 

55.  Find  the  value  of  x in  the  equation 

a4  + 2a3  — 7x 2 — 8a:=  — 12. 

Ans.  a=2, — 3,1,  or — 2 

56.  Find  the  value  of  a;  in  the  equation 

x4  — 10a3  + 35a2  — 50a: + 24  = 0. 

Ans.  a=l,  2,  3,  or  4. 

57.  Find  the  value  of  a in  the  equation 

x4  — 12a3  + 44a2  — 48a  = 9009. 

A?is.  a=13 

60.  Find  the  value  of  x in  the  equation 

a4  — 8aa3  + 8a2a;2  + 32a3a=s. 


Atis.  x—2a±^J 8a2  ± v/s+  I6n+ 


PROBLEMS. 


183 


PROBLEMS 


In  Pure  Equations  and  Affected  Quadratics  containing  but 
One  Unknown  Quantity. 


1.  Find  two  numbers  such  that  their  product  shall  be  750,  and  the 
quotient  of  the  greater  divided  by  the  less,  3^. 


Let  x represent  the  greater  of  the  two  numbers ; 

750 

then  will represent  the  less  ; and  the  Equation  will  be 

750  x2 

x-. or  = 34. 

x /o0  a 


From  this  equation  we  shall  find  2 = 50,  or  — 50.  Each  of  these 
values  will  satisfy  the  Equation  of  the  problem  ; but  only  the  positive 
one  can  be  taken  to  answer  the  conditions  of  the  problem  itself,  in 
which  the  required  numbers  are  understood  to  be  positive , as  in  the 
problems  of  common  Arithmetic.  A?is.  50,  and  15. 


2.  Find  a number  such  that  if  \ and  of  it  be  multiplied  together, 
and  the  product  divided  by  3,  the  quotient  will  be  298|-.  Ans.  224. 


3.  A mercer  bought  a piece  of  silk  for  ,£16  4s.  ; and  the  number 
of  shillings  that  he  paid  per  yard,  was  to  the  number  of  yards,  as  4 to 
9.  How  many  yards  did  he  buy  ? and  what  was  the  price  per  yard  ? 


Let  x represent  the  number  of  shillings  he  paid  per  yard  ; 

9# 

then  4 : 9 ::  x : — , the  number  of  yards. 

But  without  forming  a Proportion,  the  number  of  yards  is  readily 
known  to  be  of  the  price  per  yard. 

Ans.  27  yards,  at  12s.  per  yard. 


4.  Find  two  numbers  which  shall  be  to  each  other  as  2 to  3,  and 
the  sum  of  whose  squares  shall  be  208.  Ans.  8 and  12. 


5.  A person  bought  a quantity  of  cloth  for  $120  ; and  if  he  had 
bought  6 yards  more  for  the  same  sum,  the  price  per  yard  would  have 
been  $1  less.  What  was  the  number  of  yards?  and  the  price  per 
yard  ? Ans.  24  yards,  at  $5  per  yard. 


184 


PROBLEMS. 


6.  Divide  the  number  20  into  two  such  parts  that  the  squares 
of  these  parts  may  he  in  the  proportion  of  4 to  9.  Ans.  8,  and  12. 

7.  A merchant  bought  a quantity  of  flour  for  $100,  which  he  sold 

again  at  $5^  per  barrel,  and  in  so  doing  gained  as  much  as  each  bar- 
rel cost  him.  What  was  the  number  of  barrels  ? Ans.  20. 

8.  Divide  the  number  800  into  two  such  parts  that  the  less  divided 
by  the  greater,  may  be  to  the  greater  divided  by  the  less,  as  9 to  25. 

Let  x represent  the  less  number  : — we  shall  then  have  the  Pro- 
portion 

x 800— x 

— • : ::  9 : 25: 

800 — x x 

which  will  be  converted  into  an  Equation  by  putting  the  product  of 
the  two  extremes  equal  to  the  product  of  the  two  means. 

Ans.  300,  and  500 

9.  Two  fields  which  differ  in  quantity  by  10  acres,  were  each  sold 
for  $2800,  and  one  of  them  was  valued  at  $5  an  acre  more  than  the 
other.  What  was  the  number  of  acres  in  each  ? Ans.  70,  and  80. 

10.  A and  B started  together  on  a journey  of  150  miles.  A tra- 
veled 3 miles  an  hour  more  than  B,  and  completed  the  journey  8^ 
hours  before  him.  At  what  rate  did  each  travel  per  hour? 

Ans.  9,  and  6 miles. 

11.  A man  traveled  105  miles,  and  then  found  that  if  he  had  gone 

2 miles  less  per  hour,  he  would  have  been  6 hours  longer  on  his  jour- 
ney. At  what  rate  did  he  travel  per  hour  ? Ans.  7 miles. 

12.  A person  has  two  pieces  of  silk  which  together  contain  14 
yards.  Each  piece  is  worth  as  many  shillings  per  yard  as  there  are 
yards  in  the  piece,  and  their  whole  values  are  in  the  proportion  of  9 to 
16  ; how  many  yards  are  there  in  each  piece?  Ans.  6,  and  8 yards. 

13.  A merchant  sold  a piece  of  linen  for  $39,  and  in  so  doing 

gained  as  much  per  cent,  as  it  cost  him.  What  was  the  cost  of  the 
linen  ? Ans.  $30. 

14.  A grazier  bought  as  many  sheep  as  cost  him  $100.  After  re- 

serving 5 of  the  number,  he  sold  the  remainder  for  $135,  and  gained 
$1  a head  on  them  : how  many  sheep  did  he  buy  ? 50. 

15.  Find  two  numbers  which  shall  be  in  the  proportion  of  7 to  9. 
and  have  the  difference  of  their  squares  equal  to  128. 

Ans.  14,  and  18 

16.  An  officer  would  arrange  2400  men  in  a solid  body,  so  that 

each  rank  may  exceed  each  file  by  43  men.  How  many  must  be  placed 
in  rank  and  file  ? 75,  and  32. 


PROBLEMS. 


185 


17.  Two  partners  gained  £18  by  trade.  A’s  money  was  employed 
in  the  business  12  months  ; and  B’s,  which  was  £30,  16  months.  A 
received  for  his  capital  and  gain  £26;  what  was  the  amount  of  his 
capital  ? 

Let  x represent  A’s  capital;  then  26  — x will  be  his  gam  ; and 
since  the  gain  is  in  the  compound  ratio  of  the  capital  and  the  time  it 
was  employed,  we  have 

12^+16x30  : 12a;  ::  18  : 26— £r. 

The  first  ratio  in  this  proportion  may  be  simplified  by  dividing  the 
antecedent  and  the  consequent  by  12,  (158).  Ans.  £20. 

18  A detachment  from  an  army  was  marching  in  regular  column, 
with  5 men  more  in  depth  than  in  front ; but  upon  the  enemy’s  coming 
in  sight,  the  front  was  increased  by  845  men ; and  by  this  movement 
the  detachment  was  drawn  up  in  five  lines.  What  was  the  number 
of  men  ? Ans.  4650. 

19.  A company  at  a tavern  had  £8  15s.  to  pay,  but  before  their 

bill  was  settled,  two  of  them  went  away,  when  those  who  remained 
had  10s.  apiece  more  to  pay  than  before.  How  many  were  there  in 
the  company  at  first  ? Ans.  7. 

20.  Some  gentlemen  made  an  excursion,  and  each  one  took  the 

same  sum.  Each  gentleman  had  as  many  servants  as  there  were  gen- 
tlemen, and  the  number  of  dollars  which  each  had  was  double  the 
whole  number  of  servants  ; also  the  whole  sum  taken  with  them  was 
$3456.  What  was  the  number  of  gentlemen  1 Ans  12. 

21.  Divide  the  number  20  into  two  such  parts,  that  the  product  of 

the  whole  number  and  one  of  the  parts  shall  be  equal  to  the  square  of 
the  other.  Ans.  10  y'  5 — 10,  and  30  — 10^/5. 

22.  A laborer  dug  two  trenches,  one  of  which  was  6 yards  longer 

than  the  other,  for  £l7  16,?  , and  the  digging  of  each  cost  as  many 
shillings  per  yard  as  there  were  yards  in  its  length.  What  was  the 
length  of  each  ? Ans.  10,  and  16  yards. 

23.  There  are  two  numbers  whose  product  is  120,  and  if  2 be 
added  to  the  less,  and  subtracted  from  the  greater,  the  product  of  the 

' sum  and  the  remainder  will  also  be  120.  What  are  the  two  num- 
bers? Ans.  8 and  15. 

24.  Two  persons  lay  out  some  money  on  speculation.  A disposes 
of  his  bargain  for  £ll,  and  gains  as  much  per  cent,  as  B lays  out; 
B’s  gain  is  £36,  and  it  appears  that  A gains  4 times  as  much  per 
eent.  as  B.  What  sum  did  each  lay  out?  Ans.  A £5,  B £120. 

25.  A set  out  from  C towards  D,  and  traveled  7 miles  an  hour. 

After  he  had  gone  32  miles,  B set  out  from  D towards  C,  and  went 
each  hour  ^9  the  whole  distance ; and  after  he  had  traveled  as 
many  hours  as  he  went  miles  in  one  hour,  he  met  A.  Required  the 
distance  between  the  two  places.  Ans.  152,  or  76  miles. 


9* 


186 


SOLUTION  OF  TWO  EQUATIONS. 


Solution  of  Two  Equations — One  ok  Both  of  the  Second  ok  a 
Higher  Degree — Containing  Two  Unknown  Quantities. 

(264  ) For  the  solution  of  two  Equations,  containing  two  unknown 
quantities,  the  method  which  naturally  occurs  is, 

1.  By  elimination  between  the  given  equations  to  derive  a neio 
equation  containing  but  one  of  the  unknown  quantities,  and  thence  to 
find  the  value  of  that  quantity. 

2.  By  substituting  this  quantity  for  its  symbol  in  one  of  the  equa- 
tions containing  the  other  unknown  quantity,  to  determine  thence  the 
value  of  that  quantity. 

There  are,  however,  some  facilitating  expedients  to  be  applied,  in 
certain  cases,  to  equations  of  the  second  and  higher  degrees  : these  will 
be  exemplified  as  we  proceed. 

But  the  solution  of  two  Equations — one  or  both  of  the  second  or  a 
higher  degree — containing  two  unknown  quantities,  may  be  impossible 
by  the  method  of  quadratics , — from  the  impossibility  of  deriving  from 
them  a new  equation  containing  but  one  unknown  quantity,  which 
will  admit  of  a quadratic  solution 

EXAMPLES  AND  EXERCISES. 

1.  Find  the  values  of  x and  y in  the  equations 

2x+y=10,  and  2a;2 — a?^+3z/2=:54. 

From  the  first  equation,  we  have 

10—?/ 
nr 


„ , 2 100— 20y+y2 ),  , 10?/— ?/2 

Then  2*2=  -1 J J ' and  xy=  J J . 

4 2 

By  substituting  these  values  in  the  second  equation,  we  find 

2(  100—20 yfy2),  10  y-y2 

J — — + 3y2  = 54. 

4 o J 

The  value  of  y may  be  found  from  this  equation  ; and  by  substi- 
tuting the  value  of  y for  y in  the  first  equation,  the  value  of  x may 
readily  be  determined. 

Observe  that  the  left  hand  fraction  in  the  last  equation  may  Be  re- 
duced to  loiver  terms ; and  the  solution  of  the  equation  be  thus  some- 
what simplified. 

Ans.  x=3.  or  y= 4,  or  — J. 


EXAMPLES  AND  EXERCISES. 


187 


2.  Find  the  values  of  x and  y in  the  equations 

%-\-y  — 9,  and  x2  -\-y2  =45. 

Ans.  x—3,  oi  6 ; y—6,  or  3 

[fF=='  Whenever  x and  y may  he  interchanged  with  each  othei', 
without  changing  the  form  of  the  given  equations — as  in  the  preceding 
example — the  two  values  of  one  of  these  letters  may  be  taken,  in  re- 
verse order,  for  the-  two  values  of  the  other. 

3.  Find  the  values  of  x and  y in  the  equations 

zy —28,  and  x2-\-y2  — 65. 

Ans.  x=±l,  or  ±4;  y=-± 4,  or  ±7- 

4.  Find  the  values  of  x and  y in  the  equations 


5.  Find  the  values  of  x and  y in  the  equations 

x-\-2y=l,  and  x2 -\-3xy — y2=  23. 

Ans.  x—3,  or  15§;  y= 2,  or — 4S 

6.  Find  the  values  of  x and  y in  the  equations 


x-j-4y=14,  and  4x — 2y-\-y2  — \\. 

Ans.  x—2,  or  — 46  ; y—3,  or  15. 


2 


1 


Ans,  x=8,  or  17g  • y= 6,  or — 133. 


7.  Find  the  values  of  x and  y in  the  equations 


8.  Find  the  values  of  x and  y in  the  equations 


2 J x 


x+2 

Ans.  x—2,  or  5 ; y= 6,  or  3. 


9.  Find  the  values  of  x and  y in  the  equations 

5 (x-\-y)=13  (x — y),  and  x-\-y2=  25. 


188 


EXAMPLES  AND  EXERCISES 


Solutions  by  Means  of  an  Auxiliary  Unknown  Quantity. 

(265-)  When  the  number  of  unknoivn  factors  is  the  same  in  every 
unknown  term  of  the  two  Equations,  the  solution  will  often  be  facili- 
tated by  substituting  for  one  of  the  unknown  quantities  the  product  of 
the  other  into  a third  unk?iown  quantity. 

10.  Find  the  values  of  x and  y in  the  equations 

x2  -\-xy=5 4,  and  2xy-\-y2  =45. 

The  number  of  unknown  factors  in  each  of  the  unknown  terms  in 
these  equations,  is  two. 

If  we  assume  x to  be  equal  to  vy,  and  substitute  this  product  for  2, 
the  given  equations  will  become 

v2y2-\-vy 2 =54, 
and  2 vy2  -\-y2  =45, 

From  the  first  of  these  equations  we  have 

?/2  = . — — _ — and  from  the  second  y2  — 

v2+v,  2v+l 

Putting  these  two  values  of  y2  equal  to  each  other, 

54  _ 45 

v2 -\-v  2v-\-Y 

By  solving  this  equation  in  the  usual  manner  we  shall  find  v—2 , 
or  - — | ; then  by  substituting  these  values,  successively,  in  either  of  the 
expressions  for  y2 , we  shall  find  the  values  of  y ; and  since  X—Vy,  we 
may  also  readily  obtain  the  value  of  x. 

Ans.  a?=d=6,  or  q=9-\/  — 1 ; i/=± 3,  or  ±15-\/^l. 

If  in  the  preceding  example  an  expression  for  the  value  of  x or  y 
were  obtained  from  either  equation,  and  substituted  in  the  other,  the 
resulting  equation,  when  cleared  of  radical  signs,  would  be  of  the 
fourth  degree;  and  we  have  accordingly  found  four  values  for  each  of 
the  unknown  quantities,  (255  ) 

11.  Find  the  values  of  x and  y in  the  equations 

xy= 28,  and  x2-\-y2  = 65. 

Ans.  a:=±7,or±4;  y=±4,  or±7. 

12.  Find  the  values  of  x and  y in  the  equations 

x2-\-xy  = 12,  and  xy — 2y2  — l. 

8 1 

Ans.  x—  ±3,  or  ± — — ■>  y=±l,  or  ± — - 
\/  6 y \/  6 


SOLUTION  OF  EQUATIONS. 


189 


13.  Find  the  values  of  a and  y in  the  equations 

4a;2—  2xy=\2,  and  2y2  -{-3xy—8. 

Ans.  *,=  ±2,  or  • y=dt  1,  or±8-/i 

14.  Find  the  values  of  x and  y in  the  equations 

3 y2 — x2  =39,  and  a2 +432/  = 25  6 — 4 y2. 

Ans.  x=  ±6,  or  ± 102  ; y — ±5,  or  =p59. 


Solutions  by  Means  of  Two  Auxiliary  Unknown  Quantities. 

(266.)  When  the  unknown  quantities  are  similarly  involved  in 
each  of  the  two  Equations,  the  solution  will  sometimes  be  facilitated 
by  substituting  for  the  two  unknown  quantities  the  sum  and  difference 
of  two  other  unknown  quantities. 

15.  Find  the  values  of  x and  y in  the  equations 

x-\-y  = 12.  and  ^L+— =18. 

y x 

If  we  assume  x equal  to  v-\-z,  and  y equal  to  v — z,  we  shall  have 
x-\-y—2v  = 12  ; and  hence  v=6. 

Then  3 = 6 + 2,  and  y=  6 — z. 


Substituting  these  values  of  x and  y in  the  second  equation,  we  have 

(6+z)2+(6-Z)2_1R 

6 — z 6-} -z 

Clearing  this  equation  of  its  fractions, 

(6  + s)3  + (6— 2)3  = 18  (36—  z2). 

By  developiug  both  sides  of  this  last  equation,  and  proceeding  with 
the  solution  in  the  usual  manner,  we  shall  find  z=±2. 


Having  now  found  the  values  of  both  v and  z,  the  values  of  x and 
y are  easily  obtained. 


Ans.  a = 8or4;  y~ 4 or  8. 

16.  Find  the  values  of  x and  y in  the  equations 

a+?/=  10,  and  a;3  + ?/3  =280. 

Ans.  3=4  or  6 ; y=  6 or  4. 

17.  Find  the  values  of  x and  y in  the  equations 

x-\-y  = 11,  and  a:4+y4=2657. 

Ans.  a-=4  or  7,  y — 1 or  4 

18.  Find  the  values  of  x and  y in  the  equations 

x-\-y=  10,  and  a5  -\-ys  — 17,050. 

A';s.  a = 3 or  7,  y — 1 or  3 


190 


MISCELLANEOUS  SOLUTIONS  AND  EXERCISES. 


Miscellaneous  Solutions  and  Exercises. 

] 9.  Find  the  values  of  a and  y in  the  equations 

6,  and  x2  -\-x=\8 — y 2 — y. 

From  the  second  equation, 

x2  -\-y2  -\-x-{-y=\8. 

Adding  twice  the  first, 

x2  -\-2xy-\-y2  -\-x-\-y  = ?>t). 

This  last  equation  may  he  put  under  the  form 
(o5+7/)2+(a:+7/)  = 30  ; 

which  is  quadratic  with  reference  to  x-\-y,  and  from  which  we  may 
therefore  find  the  value  of  x-\-y. 

Ans.  35  = 2,  or  3 ; or  — 3=f  q/3  ; y — 3,  or  2 ; or — 3±  y/3. 

20.  Find  the  values  of  x and  y in  the  equations 

3"-)  ?/  = 6,  and  x2y2  +435^  = 96. 

Ans.  35  = 2,  or  4-  ; or  3 -\/ 21 ; y=4, or2  ; or3±y/21 

21.  Find  the  values  of  35  and  y in  the  equations 

x3y2=2y2,  and  835s—  t/2  = 14. 

3 2 1 L 2 

Dividing  the  first  equation  by  y2,  we  find  x3  =2y2 , or  y2  =^x3 , 

1 J 

and  by  substituting  this  value  of?/2  iu  the  second  equation,  that  equa- 
tion will  become  quadratic  with  reference  to  a:3. 

Ans.  a;  = 2744,  or  8 ; y = 9604,  or  4. 

22.  Find  the  values  of  x and  y in  the  equations 

x 2y — 25?/=  6,  and  x3y—y= 21. 

Dividing  the  first  equation  by  the  second,  we  have 

35 2 y xy  6 

x3  y y — 21 

This  equation  will  be  simplified  by  reducing  each  of  its  two  frac- 
tional members  to  its  lowest  terms.  1 

Ans.  x=2,  or  2 ! 2/ = 3,  or  — 24. 

23.  Find  the  values  of  x and  y in  the  equations 

%3jrzy2  =39,  and  x2y-{-y3  = 26. 

Ans.  35=3  ; y= 2. 


MISCELLANEOUS  SOLUTIONS  AND  EXERCISES. 


191 


24.  Find  the  values  of  x and  y in  the  equations 

x — y— 4,  and  a:3 — t/3=316. 

Dividing  each  side  of  the  second  equation  by  the  corresponding 
side  of  the  first,  we  shall  find,  x2  -\-xy-\-y2  =79. 

Squaring  the  first  equation,  and  subtracting,  we  have  3xy=63 

Ans  x=7,  or  — 3;  y~ 3,  or  — 7 

25.  Find  the  values  of  x and  y in  the  equations 

x2 — y2 —5,  and  a:4 — r/4=65. 

Ans.  x—  ±3,  y~± 2 

26.  Find  the  values  of  x and  y in  the  equations 

x-\-y— 60,  and  2 (a2 -\-y2)  — 5xy. 

The  second  equation  may  be  put  under  the  form  x2  -\-y 2 — 2^xy  — (S  , 
and  the  solution  will  be  facilitated  by  subtracting  this  from  the  square 
of  the  first  equation.  Ans.  x=40,  or  20  ; y^=  20,  or  40. 

27.  Find  the  values  of  x and  y in  the  equations 

x2  y2 

xy  = 8,  and \-  — = 9. 

y x 

Multiplying  the  two  equations  together,  we  find  x 3 - \-y 3 =72  ; and, 
multiplying  this  equation  by  x3,  we  have  x6  -{-x3  y3  = 7 2x3 . 

From  the  first  equation,  x3  y3  =83  =512  ; and  if  this  number  be 
substituted  in  the  preceding  equation  we  shall  have  a quadratic  with 
reference  to  x3.  Ans.  x=4,  or  2;  y=. 2,  or  4. 


28.  Find  the  values  of  x and  y in  the  equations 

xij—  25,  and  x3  +y3  = lOxy. 

The  second  equation  will  be  reduced  to  the  same  form  as  the 
second  in  the  preceding  example,  by  dividing  it  by  xy. 

Ans.  X—5  . y—5. 

29.  Find  the  values  of  x and  y in  the  equations 

x2 — y2 — (_c+r/)  = 8,  and  ( x — r/)2(x+?/)  = 32. 


Dividing  each  equation  by  x-\ -y,  we  have 

8 NO  32 

x — v — 1= , and  (x — 1/)2= • 

x+y  x+y 

Transposing  — 1,  and  squaring,  we  obtain 

, / 8 \2  32 

{X~y)  -C+y)+1)  =x+y] 

from  which  will  result  a quadratic  with  reference  to  x-\~y. 

Ans.  x—5  ; y= 3. 


192  MISCELLANEOUS  SOLUTIONS  AND  EXERCISES. 

30.  Find  the  values  of  x and  y in  the  equations 

x-\--\/xy-{-y—l , and  x2  +xy+y2  =21. 

Dividing  the  second  equation  by  the  first,  we  have 
x — -\/xy+y~3. 

W e shall  now  obtain  two  equations  of  simpler  forms  by  adding  the 
third  equation  to  the  first,  and  subtracting  it  from  the  first. 

, Ans.  x=\,  or  4 ; y— 4,  or  1 


31.  Find  the  values  of  x and  y in  the  equations 

x-\-y  , — z — 2a cy  12 

-g-=V*y+4,  and  Vxy=^^+-j- 

From  the  first  equation,  by  transposition, 

— x4 -y 
*xym2 — 4‘ 

Clearing  this  equation  of  its  fraction,  and  squaring, 

. 4xy=(x+y)2 — 1 6 (z+?/)  + 64. 

By  equating  the  two  values  of  sfxy,  from  the  second  and  third 
equations, 

2xy  12  _x+y 
x+y + 5 ~ 2 


or 


2 xy  x + y 
x+y~  2 


32 

" T" 


Clearing  this  equation  of  its  fractions, 

64 

4,xy={x+y)2 — ~(x+y). 

We  shall  now  obtain  a simple  equation  by  subtracting  this  last 
equation  from  the  fourth  equation. 

Ans.  x = 2,  or  18;  ^ = 1 8,  or  2. 


32.  Find  the  values  of  x and  y from  the  equation  and  proportion 

xy2 — x—3  ; 

x2  ?/4 — x 2 : x2+x2  y2+x2yi  : : 5:7. 

In  any  proportion  the  difference  of  the  first  and  second  terms  is  to 
the  first,  as  the  difference  of  the  third  and  fourth  is  to  the  third,  (160). 
Hence  from  the  given  proportion,  we  shall  find 
2 x2+x2y2  : x2yi — x2  . : 2:5. 

Dividing  the  first  antecedent  and  consequent  by  x2,  (15S) 

2 +y2  : y*  — 1 : : 2:5. 


MISCELLANEOUS  SOLUTIONS  AN!  EXERCISES. 


193 


33.  Find  the  values  of  x and  y in  the  equations 

x2  -\-xy-\-y2  = 13,  and  x*  -\-x2y2  -\-y*  =91. 

Dividing  the  second  equation  by  the'first, 
x2  — xy-\-y2  — 7. 

Adding  the  third  equation  to  the  first,  and  dividing  the  result  by 
i! : and  also  subtracting  the  third  from  the  first, 
x2  - \-y 2 = 10  ; 
and  2xy=&. 

The  solution  now  proceeds  by  adding  the  latter  of  these  two  equa- 
tions to  the  former,  and  also  subtracting  the  latter  from  the  former,  and 
extracting  the  square  roots  of  the  resulting  equations. 

Ans.  a;=:±3;  y—  ± 1. 

34.  Find  the  values  of  x and  y in  the  equations 

x2y—x2y2 — x2,  and  x2y2  -\-x2  =x3y3 — x3 

Dividing  each  equation  by  x2,  we  have 

y—y2  — 1,  and  y2  -\-\—xy3 — x. 

The  value  of  y is  to  be  found  from  the  first  of  these  equations,  and 
substituted  in  the  second.  Ans.  x=ly'o  ; y=\ ^/5. 

35.  Find  the  values  of  x and  y in  the  equations 

xy—X 2 — y2,  and  x2  -j-y2  =z3 — y3. 

If  we  assume  y to  be  equal  to  xv,  and  substitute  this  product  for 
y in  the  two  equations,  (264,)  we  shall  have 

x2v—x 2 — x2v2,  and  x2  -\rx2v2  =a?3 — x3v3. 

These  equations  may  be  solved  in  the  same  manner  as  those  in  the 
preceding  example.  Or  the  value  of  i may  be  found,  in  terms  of  y, 
from  the  first  of  the  given  equations,  and  substituted  in  the  second. 

Thu»3:=I±V/¥=I±^- 

2 v 4 2 2V 

Then  for  the  second  equation, 

Ans.  x—\{5±  y—±\^5. 


MISCELLANEOUS  SOLUTIONS  AND  EXERCISES. 


i 4 

36.  Find  the  values  of  x and  y in  the  equations 
x3 — x2y — xy2  -\-y3  =576,  and  a;3  -\-x2y-\-xy2  -\-y3  =2336 

Subtracting  the  first  equation  from  the  second, 

2x2y+2xy2  — \l&Q. 

Adding  this  to  the  second  equation,  we  obtain 
23-f-3x2t/-|-3a:?/2-|-t/3=4096. 

By  extracting  the  cube  root  of  this  equation,  we  shall  fun  the 
value  of  x-\-y,  which  may  be  substituted  for  x-\-y  in  the  third  equ.  tion. 

-Arcs.  2=11;  y~5 

37.  Find  the  values  of  x and  y in  the  equations 

au/  = 320,  and  x3  —y3  =61  (2 — y)3. 

Dividing  the  second  equation  by  x — y,  we  have 
x2  +xy-\-y2  =61  (2—  y) 2. 

By  converting  this  equation  into  a proportion,  (153,) 
x2  -{-  xy-\-y2  : ( x — y)2  : : 61  : 1. 

The  solution  now  proceeds  by  developing  the  term  (x — y)2, — sub- 
tracting each  consequent  from  its  antecedent,  and  forming  a proportion 
of  the  consequents  and  remainders,  Ac,  (160). 

Ans.  2=20  ; y=  16. 

38.  Find  the  values  of  x,  y,  and  z,  in  the  equations, 

x2 -\-y2 -\-xy=.Zl  ; x2 -\-z2 -\-xz— 49;  y2  +z2  +2/2  = 61. 

Subtracting  the  first  equation  from  the  second,  and  decomposing, 
(z—y)(z+y)  + {z—?/)x,  or  (z+y+x)(z-y)  = 12  ; 

12 

from  which  z-\-y-\-x  — — - — • 
z y 

Proceeding  in  like  manner  with  the  second  and  third  equations, 

12 

we  shall  find  y-\-x-{-z= ‘ 

y—x 

Hence  the  right  hand  members  of  the  last  two  equations  are  equal 

to  each  other,  (113 1);  and  since  the  numerators  are  the  same,  we 

have 

z — y—y — x,  from  which  2 y=x-\-z. 

By  substituting  2 y for  x-\-z  in  the  sixth  equation,  we  shall  fiud 
y2  — yx—\. 

The  value  of  2 from  this  equation,  is  to  be  substituted  in  the  first 
equation.  Ans.  2 = 3;  y~ 4;  c = <3. 


PROBLEMS. 


195 


PROBLEMS 


In  Quadratic  Equations  of  One  or  More  TJnknowr. 
Quantities. 

1.  The  area  of  a rectangular  lot  of  ground  is  384  square  rods,  and 
its  length  is  to  its  breadth  as  3 is  to  2.  Required  the  length  and 
breadth  of  the  lot. 


The  area , in  square  measure,  of  a rectangle,  is  expressed  by  the 
product  of  the  number  of  linear  units  in  its  length  X the  number  of 
linear  units  in  its  breadth. 

The  length  and  breadth  must  be  taken  in  the  same  denomination 
in  multiplying  : the  area  will  be  found  in  the  corresponding  denomina- 
tion of  square  measure. 


Let  x represent  the  length,  and  y the  breadth  of  the  lot ; then  by 
the  conditions  of  the  problem, 

xy  — 384, 

and  x : y \ . 3:2. 

2x 

Or,  if  x represent  the  length,  - will  represent  the  breadth,  and 

o 

we  shall  then  have 


2x2 


= 384. 


Ans.  24,  and  16  rods. 


2.  The  length  of  a rectangular  garden  exceeds  its  breadth  by  6 

rods,  and  its  area  is  216  square  rods.  What  are  the  length  and 
breadth  of  the  garden  ? Ans.  18,  and  12  rods. 

3.  Find  two  numbers  whose  sum  shall  be  24,  and  whose  product 

shall  be  equal  to  35  times  their  difference.  Ans.  14  and  10. 

4.  Divide  a line  20  inches  in  length  into  two  such  parts  that  the 
rectangle  or  product  of  the  whole  line  and  one  of  the  parts  shall  be 
equal  to  the  square  of  the  other  part. 

Ans.  10y/5— 10,  and  30— lOV^. 

5.  Find  the  dimensions  of  a rectangular  field,  so  that  its  length 
shall  be  equal  to  twice  its  breadth,  and  its  area  800  square  rods. 

Ans.  40,  and  20  rods. 

6.  The  sum  of  the  two  digits  of  a certain  number  is  10,  and  il 

their  product  be  increased  by  40,  the  digits  will  be  reversed.  What 
is  the  number  ? Ans.  46. 


196 


PROBLEMS. 


7.  The  sum  of  two  fractions  is  li,  and  the  sum  of  their  recip  <jcah 

is  3^  ; what  are  the  two  fractions  1 Am.  \ and 

8.  The  perimeter , or  sum  of  the  four  sides  of  a rectangle,  is  112 

rods,  and  its  area  is  720  square  rods.  What  are  the  length  and  breadth 
of  the  rectangle  ? Ans.  36,  and  20  rods. 

9.  Divide  the  number  60  into  two  such  parts  that  their  product 

shall  be  to  the  sum  of  their  squares  as  2 to  5.  Ans.  20  and  40 

10.  A merchant  bought  a piece  of  cloth  for  $120,  and  after  cutting 

off  4 yards,  sold  the  remainder  for  what  the  whole  cost  him — by  which 
he  made  $1  a yard  on  what  he  sold.  How  many  yards  did  the  piece 
contain  ? Ans.  24. 

11.  Divide  the  number  100  into  two  such  parts  that  the  difference 

oftheir  square  roots  shall  be  2.  Am.  64,  and  36. 

12.  A garden  which  is  20  rods  square  is  surrounded  by  a walk 

whose  area  is  equal  to  of  the  area  of  the  garden  itself.  What  is  the 
breadth  of  the  walk  ? g^/o—10  rods. 

13.  The  sum  of  the  squares  of  two  numbers  is  325,  and  the  dif- 
ference of  their  squares  is  125.  What  are  the  numbers  ? 

Am.  15  and  10. 

14.  The  area  of  a rectangular  court-yard  is  875  square  rods,  and  if 
its  length  and  breadth  were  each  increased  by  5 rods,  its  area  would 
then  be  1200  square  rods.  What  are  the  dimensions  of  the  yard  ? 

Am.  35  and  25  rods. 

15.  The  difference  of  two  numbers  is  4,  and  the  difference  of  their 

cubes  is  448.  What  are  the  two  numbers  ? Am.  8 and  4. 

16.  A grocer  sold  80  pounds  of  mace  and  100  pounds  of  cloves  for 
£65,  and  finds  that  he  has  sold  60  more  of  cloves  for  £20  than  of  mace 
for  £10.  What  was  the  price  of  each  per  pound. 

Ans.  10s.  and  5s. 

17.  The  fore- wheel  of  a carriage  makes  6 revolutions  more  than  i 
the  hind  wheel  in  going  120  yards;  but  if  the  circurnferenee  of  each  i 
be  increased  1 yard,  it  will  make  only  four  revolutions  more  in  going  | 
the  same  distance.  What  is  the  circumference  of  each  wheel  ? 

Ans.  4,  and  5 yards. 

18.  Find  four  numbers  in  arithmetical  progression,  such,  that  the 
product  of  the  two  extremes  shall  be  45,  and  the  product  of  the  two 
means  77. 

L.et  a:  be  the  first  term,  and  y the  common  difference  of  the  terms  ; I 
then  the  numbers  will  be 

x,  x-\-y,  x-\-2 y,  £+3 y ; (175). 

and  by  the  conditions  of  the  problem  we  shall  have 

a;2-)- 3a:?/ =45  ; and  x2  + 3.W/+2?/2  =77. 

H??s.  3,  7,  11,  and  16. 


PROBLEMS 


19? 


19.  A farmer  has  a field  16  rods  long  and  12  rods  wide,  which  he 

wishes  to  enlarge  so  that  it  may  contain  just  twice  as  much  area, 
without  altering  the  proportion  of  the  sides.  What  will  be  the  dimen- 
sions of  the  field  when  thus  enlarged?  i6-y/2  ; and 

20.  Find  three  numbers,  such,  that  the  difference  of  the  first  and 
second  shall  be  two  less  than  the  difference  of  the  second  and  third, 
their  sum  17,  and  the  sum  of  their  squares  115. 

The  solution  will  be  facilitated  by  assuming  x to  represent  the 
second  number,  and  y the  difference  of  the  first  and  second. 

Ans.  3,  5,  and  9. 

21.  There  are  two  square  gardens  which  together  contain  1025 

square  rods,  and  a side  of  the  one  exceeds  a side  of  the  other  by  5 rods. 
What  are  the  sides  of  the  two  gardens?  Ans.  20,  and  25  rods. 

22.  Find  two  numbers,  such,  that  their  sum,  their  product,  and 
the  difference  of  their  squares  shall  all  be  equal  to  one  another. 

Take  x-\-y  to  represent  the  greater,  and  x — y the  less  number. 

Ans.  |± and  \±y/\. 

23.  A merchant  received  $12  for  a quantity  of  linen,  and  an  equal 
sum,  at  50  cents  less  per  yard,  for  a quantity  of  calico,  which  exceeded 
the  quantity  of  linen  by  32  yards.  What  was  the  quantity  of  each  ? 

Ans.  16,  and  48  yards. 

24.  Find  two  numbers  whose  sum  multiplied  by  the  greater  shall 
be  equal  to  192,  and  whose  difference  multiplied  by  the  less  shall  be 
equal  to  32. 

The  solution  will  be  facilitated  by  taking  x to  represent  one  of  the 
required  numbers,  and  xy  the  other.  Ans.  12,  and  4. 


25.  Three  merchants  gained  $1444  ; of  which  their  respective 

shares  were  such  that  B’s,  added  to  the  square  root  of  A’s,  made  $920: 
but  if  added  to  the  square  root  of  C’s  it  made  $912.  Wrhat  was  the 
share  of  each  ? Ans.  $400,  $900,  and  $144. 

26.  The  sum  of  three  numbers  in  harmonical  progression  is  13> 
and  the  product  of  the  two  extremes  is  18.  What  are  the  numbers  ? 


If  x and  y represent  the  two  extremes,  the  mean  term  will  be 

2xv  (184). 

x-\-y  ' Ans.  6,  4 and  3. 


27.  There  is  a rectangular  field  whose  length  is  to  its  breadth  as  4 
to  3.  A part  of  this  field,  which  is  equal  to  i of  the  whole,  being  in 
meadow,  there  remain  for  ploughing  1296  square  rods.  WThat  are  the 
dimensions  of  the  field  ? Ans.  48,  and  36  rods. 


198 


PROBLEMS. 


28.  The  sum  of  three  numbers  in  geometrical  progression-  is  21,  ; 
and  the  sum  of  their  squares  is  189.  What  are  the  numbers  ? 

If  x and  y represent  the  two  extremes,  the  mean  term  will  be 

'\fxy,  (189).  Ans.  3,  6,  and  12. 

29.  A and  B set  out  from  two  places  which  are  distant  110  miles, 
and  traveled  towards  each  other.  A went  five  miles  an  hour  ; and  the 
number  of  hours  in  which  they  met  was-  greater,  by  four,  than  the 
number  of  miles  B went  per  hour.  What  was  B s rate  of  traveling  ? 

Ans.  6 miles  per  hour. 

30.  The  arithmetical  mean  between  two  numbers  exceeds  the  geo-  fi 

metrical  mean  by  13,  and  the  geometrical  mean  exceeds  the  harmoni-  | 
cal  mean  by  12.  What  are  the  numbers?  Ans.  234  and  104.  |j 

31.  Three  merchants  made  a joint  stock,  by  which  they  gained  a 
sum  less  than  that  stock  by  $80.  A’s  share  of  the  gain  was  $60,  and  i 
his  contribution  to  the  stock  was  $17  more  than  B's ; also  B and  0 : 
together  contributed  $325.  How  much  did  each  contribute  ? 

Ans.  $75,  $58,  and  $267  | 

32.  Of  three  numbers  in  geometrical  progression  the  greatest  ex- 
ceeds the  least  by  15,  and  the  difference  of  the  squares  of  the  greatest 
and  the  least  is  to  the  sum  of  the  squares  of  the  three  numbers  as  5 to  7. 
W?hat  are  the  numbers  ? 

Assume  x to  represent  the  first  term,  ond  y the  ratio  of  the  pro- 
gression. Ans.  5,  10,  and  20, 

33.  Two  persons  set  out  from  different  places,  and  traveled  towards  i 

each  other.  On  meeting,  it  appeared  that  A had  traveled  24  miles  I 
more  than  B,  and  that  A could  have  gone  B’s  journey  in  8 days,  while  |i 
B would  have  been  18  days  in  performing  A’s  journey.  1/V  hat  distance 
was  traveled  by  each  ? Ans.  72,  and  48  miles.  Jj 

34.  The  joint  stock  of  two  partners  was  $416.  A’s  money  was  in  j 

the  business  9 months,  and  B’s  6 months.  When  they  shared  stock  II 
and  gain,  the  first  received  $22S,  and  the  second  $252  ; what  was  each  I j 
man’s  amount  of  stock?  Ans.  A s $192,  B s $224.  ‘J 

35.  The  sum  of  $700  was  divided  among  four  persons,  A,  B.  C,  and  j I 
D,  whose  shares  were  in  geometrical  progression  ; and  the  difference  It 
between  the  greatest  and  the  least  was  to  the  difference  between  the  .1 
two  means  as  37  to  12.  What  were  the  several  shares  ? 

Ans.  $108,  $144,  $192,  and  $256.  ■ } 


CUBIC  AND  HIGHER  EQUATIONS. 


199 


Solution  of  Affected  Cubic  and  Higher  Equations. 

Various  methods  have  been  devised  for  the  solution  of  Affected 
Equations  of  the  third  and  higher  degrees.  Some  of  these  methods 
are  very  prolix, — while  others  are  of  limited  application  ; we  shall  ex- 
plain those  which  are  the  most  useful  in  a practical  point  of  view, 
without  attempting  a full  exposition  of  this  subject. 

Under  the  head  of  General  Properties  of  Equations,  we  have  al- 
ready noticed  the  divisors,  (253),  and  the  number  of  roots,  (255)  of 
equations  ; we  here  present  the 

_ 

General  Law  of  the  Coefficients  of  Equations. 

(267)  When  the  terms  of  any  complete  Equation  containing  but  one 
unknown  quantity  x,  are  all  arranged,  according  to  the  descending 

I powers  of  x,  hi  the  first  member — with  the  known  or  absolute  term  for 
the  last  term — and  the  coefficient  of  the  first  term  is  unity ; then 

1.  The  coefficient  of  the  second  term  is  equal  to  the  sum  of  all  the 
roots  of  the  equation,  with  their  signs  changed. 

2.  The  co-efficient  of  the  third  term  is  equal  to  the  sum  of  the  pro- 
ducts of  all  the  roots  combined  two  and  two,  with  their  signs  changed, 

&c. 

3.  The  known  or  absolute  term  is  equal  to  the  product  of  all  the 
roots,  with  their  signs  changed. 

To  demonstrate  these  principles  with  reference  to  a Cubic  Equa- 
tion, let  the  three  roots  be  denoted  by  a,  b,  and  — c ; then  x — a'  x — b, 
and  x 4-  care  th  e divisors  of  the  equation,  and  the  equation  may  be  ac- 
cordingly resolved  into 

(x — a)  ( x — b)  (.r+c)  =0,  (253). 

By  performing  the  multiplication  which  is  here  indicated,  and  de- 
composing the  terms  containing  the  like  powers  of  x in  the  product, 
we  find 

x3  + (c — a — b)  x2-\-(ab — ac — be)  x-fabcz^Q. 

In  this  equation  the  coefficient  of  x2  is  the  sum  of  the  roots  a,  b, 
and  — c,  with  their  signs  changed  ; the  coefficient  of  a;  is  the  sum  ol 
the  products  of  the  roots  combined  two  pnd  two,  with  their  sign? 
changed  ; and  the  known  or  absolute  term  abc  is  the  product  of  all 
the  roots,  with  their  signs  changed. 

The  same  principles  may  be  demonstrated,  in  like  manner,  in  refer- 
ence to  an  Equation  of  the  second,  or  of  any  of  the  higher  degrees. 


200 


CUBIC  AND  HIGHER  EQUATIONS. 


An  application  of  these  principles  may  be  made  to  the  equation 
z2  3x — 10  = (a; — 2)  (z-j-5)  = 0,  whose  roots  are  2,  and  — 5 ; or  to 
c3  — 19a:-)- 30 ~(x — 2)  (a-(-5)  ( x — 3)  = 0,  whose  roots  are  2,  — 5,  & 3 

In  the  second  equation  it  will  be  observed  that  the  second  term  \ 
containing  x 2 , is  wanting,  since  its  coefficient,  that  is,  the  sum  of  th<  i 
roots  2,  —5,  and  3,  is  0 ; and  that  19a:  therefore  corresponds  to  the 
third  term,  (267. ..2). 

' I 

Determination  of  the  Integral  Roots  of  Equations. 

(268  ) If  an  equation  containing  hut  one  unknown  quantity  a:,  wit! 
all  its  terms  transposed  to  one  side,  be  divisible  by  a:-)-  or  — any  num 
her,  that  number,  with  a contrary  sign,  will  be  a root  of  the  equation 
(254). 

The  trial  numbers  to  be  used  in  this  division,  are  the  factors  or  ch 
visors  of  the  known  term  of  the  equation,  since  that  term  is  equal  to 
the  product  of  all  the  roots  of  the  equation,  (267. ..3). 

When  an  equation  has  any  integral  roots , such  roots  may  be  readily 
determined  by  an  application  of  these  principles. 


EXAMPLE. 

To  find  the  values  of  x in  the  equation 

a:3  + 3x’2  — 4.r=12. 

The  divisors  of  the  known  term  12  are  1,  2,  3,  4,  6,  and  12;  and 
it  will  he  found,  on  trial,  that  the  equation 

x3-)-3x2 — 4a: — 12  = 0, 

is  divisible  by  a:  — 2,  x-\-2,  and  cc -4- 3 ; hence  the  values  of  x,  or  roots 
of  the  equation,  are  2,  — 2,  and  — 3. 

After  any  one  of  the  three  roots  has  been  determined,  the  two  re- 
maining ones  may  be  obtained  directly  from  the  quadratic  equation 
which  results  from  dividing  the  given  equation  by  X-\-  or  — the  root 
already  found. 

x — 2 ) z3  -\-3x2  —4a: — 12  ( x2  +5a:-)-6. 

By  dividing  the  given  equation  by  x — 2,  we  thus  find 
r-2-|-5x-(-6  = 0, 

or  x2+5a:=: — 6,  which  gives  x=  — 2 or  —3 
By  this  method  the  last  two  routs  are  found  the  same  as  before. 


CUBIC  AND  HIGHER  EQUATIONS. 


201 


Solution  of  Equations  by  Approximation. 

(2G9.)  The  following  method  of  solution  may  he  applied  to  an 
Equation  of  any  degree — even  to  one  in  which  the  unknown  quantity 
is  left,  without  rationalization,  in  a surd  expression. 

1 . By  trial  find  two  numbers — differing  by  a unit  or  less — which 
being  substituted  for  x in  the  given  Equation,  will  produce  results, 
the  one  less  and  the  other  greater  than  the  knoivn  term  of  the  equa- 
tion ; then, 

The  difference  between  the  tivo  results, 

Is  to  the  difference  between  the  two  assumed  numbers, 

A.s  the  difference  between  either  result  and  the  knoivn  term, 
Is  to  the  correction,  nearly,  required  in  the  corresponding  as- 
sumed number. 

2.  Take  the  corrected  root  thus  obtained  for  one  of  two  numbers  to 
be  substituted  for  x,  and  find,  and  apply,  a correction  as  before. 

We  shall  thus  obtain  a nearer  value  of  the  unknown  quantity  ; 
and  the  approximation  may  be  carried,  in  like  manner,  to  any  required 
exactness.  _ 

EXAMPLE. 

To  find  an  approximate  value  of  x in  the  equation. 

a3+a;2+a:=100. 

First,  It  will  be  found  that  x is  more  than  4,  and  less  than  5.  Sub 
stituting  these  numbers  for  x,  we  have 


64.... 

....125 

16... 

....  25 

4.. . 

. . . . x . . . . 

...  5 

84 

155 

The  difference  between  the  two  results  is  155  — 84  = 71  ; and  the 
difference  between  the  less  result  and  the  known  term  100  is  16. 
Then  71  : 1 : : 16  : the  correction  . 225. 

This  correction,  added  to  the  less  assumed  number,  gives  4.225  for 
an  approximate  value  of  x. 

Secondly,  By  substituting  4.2  and  4.3  for  x,  we  have 


74.088... 

. . . a;3  . . . 

...79.507 

17.64  ... 

...18,49 

4.2  ... 

. . . x . . . 

...  4.3 

95.928 


10 


102.297 


202 


CUBIC  AND  HIGHER  EQUATIONS. 


Forming  a proportion  between  the  difference  of  these  two  results, 
and  the  difference  between  greater  result  and  the  known  term  100, 
6.369  : ,1  ::  2.297  : the  correction  .036. 

By  subtracting  this  correction  from  the  greater  assumed  number 
4 3,  we  have  4.264  for  a nearer  value  of  x. 

For  the  next  approximation  we  should  take  4.264  and  4,265  to 
to  be  substituted  for  x in  the  given  Equation.  The  value  of  x would 
then  be  found  to  be  4.2644299  very  nearly. 

In  the  Proportion  for  finding  the  correction,  it  is  best  to  employ  the 
less  error  in  the  results  of  the  substitution. 

Thus  in  the  first  substitution,  in  this  Example,  the  error  in  the  less 
result  84  is  (100  — 84)=:  16,  and  this  being  less  than  the  error  in  the 
155,  we  employ  16  in  the  first  proportion. 

But  in  the  second  substitution,  the  error  in  the  greater  result 
102.297  is  less  than  the  error  in  the  95.928,  and  we  accordingly  use 
(102.297  — 100)  = 2. 297  in  the  second  proportion. 

Each  approximative  solution  will  generally  double  the  number  of 
true  figures  in  the  root.  Thus  in  the  preceding  Example  we  found 
by  trial  that  4 is  the  first  figure  in  the  root,  and  the  first  solution  gives  i 
4.2  for  the  first  two  correct  figures  ; the  next  solution  gives  4.264  ; j 
and  the  number  of  figures  will  again  be  doubled  by  a third  solution. 
This  property  determines  the  number  of  figures  which  need  be  found 
in  the  successive  corrections  of  the  assumed  numbers. 

To  find  the  other  Roots  of  the  given  Equation,  we  would  divide 
a,3  + a2+a; — 100  = 0 by  Z— 4,2644,  &c.  = 0,  (253). 

"We  should  thus  obtain  a quadratic  equation,  from  which  the  other 
two  values  of  x might  be  determined,  according  to  the  usual  method. 

"When  all  the  Boots  of  an  Equation  have  been  found,  we  may 
verify  them  by  the  property  that,  with  their  signs  changed,  their  sum 
must  be  equal  to  the  coefficient  of  the  second  term  of  the  equation,  j 
(267. ..1). 

Thus  the  sum  of  the  three  roots  of  the  equation  in  the  preceding 
Example,  with  their  signs  changed,  would  be  unity,  which  is  the  co- 
efficient of  the  second  term  a2. 


CUBIC  AND  HIGHER  EQUATIONS. 


203 


EXERCISES 

On  Affected  Cubic  and  Biquadratic  Equations. 


I .  Find  the  values  of  x in  the  equation 

x3  — 6a;2 + 11*  = 6,  (268). 


2.  Find  the  values  of  x in  the  equation 

x3  — 9x2  + 26x  = 24. 

3.  Find  the  values  of  x in  the  equation 

x3  — 3x2 — 6x=  — 8. 


Ans.  x=l,  2,  or  3. 
Ans.  x=2,  3,  or  4. 
Ans.  x=l,  4,  or  — 2. 


4 Find  the  values  of  x in  the  equation 

2x3— e^2— Sx=— 24. 

Ans.  x=2,  3,  or  — 2. 

5.  Find  the  values  of  x in  the  equation 

x4+2x3  — 13x2  — t4x+24  = 0. 

Ans.  1,  —2,  3,  or  —4 


6.  Find  an  approximate  value  of  x in  thP  t_  -pation 

x3  + 10x2  + 5x=260,  (26^). 

7.  Find  an  approximate  value  of  x in  the  equ  Mon 

x3  — 15x2  + 63x=50. 

8.  Find  an  approximate  value  of  x in  the  eqv  tion 

x3  — I7x2  -t-54x  = 350. 

9.  Find  an  approximate  value  of  x in  the  eqi?’  ton 

x4  — 3x2  — 75x=  10000. 

0.  Find  an  approximate  value  of  x in  the  equation 
2x4  — 16x3  + 40x2  — 30x+ 1 = P. 

1.  Find  an  approximate  value  of  x in  the  equation 
Qx2 — 15)2+x^/x  = 90. 


Ans.  £=4.117’. 


Ans.  x=  1.028.’ 


aL/is.  x= 14.95’. 


Ans.  £=10.23’. 


An*  an-'  1.284’ 


Anz.  x=laJ'’ 


K 


204 


GENERAL  METHOD  OF  ELIMINATION. 


Elimination  by  the  Method  of  Common  Divisor. 

(269.)  The  following  is  a general  method  of  Elimination,  and,  for  ■ 
Equations  of  the  higher  degrees , it  will  sometimes  be  found  preferable 
to  any  other. 

Transpose  all  the  terms  of  the  two  Equations  to  one  side  ; then  di- 
vide one  into  the  other,  and  the  remainder  into  the  divisor , and  so 
on,  as  in  finding  the  (freatest  Common  Measure,  (66,)  until  one  of  the 
two  unknown  quantities  is  eliminated  from  the  remainder ; and  put 
this  remainder  =0. 

To  eliminate  x from  the  equations 

x2  -{-xy—\0,  and  xy  + 2y2  — 24. 

xy+2y2  — 24 
x 

x2-\-xy — 10  j x2y-\-2xy22\ — xly 
f x2y-\-xy2  — lOy 

xy2 — 24x+10 y,  or  x{y2  — 24)  + 10_y. 

x2  +xy — 10 
y2—  24 

xiy2  — 24)+  \0y  1 a,2(y2—  24)  + a;(y3  — 24_y)  — 10y2 +240(.r+y 
) x2[y2 — 24)+10 xy 

x(y 3 — 24w)  — 1 Oxy—  1 0y2  + 240 
x{y3  — 24  y)  + 10  y2 

— 10  xy — 20y2  +240 

x{y2 -24)  + 107/ 

y 

a?/+2y2  — 24  ) x[y 3 — 24 y)  + 10 w2  {y2  — 24 

/ x(y3  — 24?/)  + 2y4  — 48y2  — 24y2  + 576 

— 2y4  + S2y2— 576~0. 

In  the  remainder  — lOxy— 20?/2  + 240,  we  cancel  the  factor  10 
and  change  the  signs,  for  the  next  divisor.  In  dividing  into  this  di 
visor,  we  take  the  binomial  y2  — 24  for  the  quotient,  and  multiply  the 
divisor  by  this  binomial. 

The  first  remainder  is  equal  to  0,  because  the  divisor  and  dividenc 
are  each  equal  to  0 ; and  it  follows  hence  that  each  subsequent  re  \ 
mainder  is  equal  to  0. 

The  operation  will  be  much  more  simple  if  we  divide  the  firs 
equation  by  the  second  : the  result  will  be  the  same. 


205 


CHAPTER  XI. 

«w 

GENERAL  DESCRIPTION  OF  PROBLEMS. INEQUATIONS. 

Miscellaneous  Problems. 

1.  Determinate  Problems. 

(270.)  A Determinate  Problem  is  one  in  which  the  given  condi- 
tions determine  the  values  of  the  unknown  or  required  quantities. 

A Determinate  Problem  is  represented  by  as  many  independent 
equations  as  there  are  different  conditions  to  be  expressed,  or  unknown 
quantities  to  he  determined,  (120.) 

All  the  Problems  which  have  hitherto  been  proposed  in  this  work, 
are  determinate  j and  no  example  of  this  kind  need  be  here  given. 

2.  Indeterminate  Problems. 

(271.)  An  Indeterminate  Problem  is  one  in  which  the  given  condi- 
tions do  not  determine  the  values  of  the  required  quantities, — admitting 
either  of  an  unlimited  number  of  values  to  those  quantities,  or  else  of 
a variety  of  values,  within  certain  limits. 

An  Indeterminate  Problem  is  represented  either  by  a less  number  of 
independent  Equations  than  there  are  unknown  quantities  to  be  deter- 
mined, or  by  an  identical  equation. 

We  give  an  example  of  each  of  these  forms  of  indeterminateness. 

Example  I. 

To  find  three  numbers  such  that  the  first  shall  be  5 less  than  the 
second,  and  the  sum  of  the  second  and  third  shall  be  12. 

This  Problem  contains  but  two  conditions ; and  if  we  represent  the 
three  required  numbers  by  x,  y , and  z,  we  shall  have  only  the  two 
Equation^ 

y—x—5; 
y+z= 12. 

By  subtracting  the  first  equation  from  the  second,  we  have 


206 


6ENERAL  DESCRIPTION  OF  PROBLEMS. 


X + Z = 7. 

This  equation  will  admit  of  an  unlimited  number  of  values  of  x 
and  z ; for  we  may  assume  any  value  whatever  for  one  of  the  letters, 
as  x,  and  determine  thence  the  corresponding  value  of  z. 

Thus  if  x—\,  Z— 6|  : if  x=\,  z— 6J;  if  *=|,  z= 6\,  &c.; 
and  ifom  the  values  of  x or  z,  we  might  obtain  the  corresponding 
values  of  y from  one  of  the  given  equations. 

If,  however,  the  required  numbers  were  limited  to  integral  values, 
the  third  equation  would  be  satisfied  only  by 

x=l,  2,  3,  4,  5,  or  6,  and  z = 6,  5,  4,  3,  2,  or  1. 

In  the  first  equation  y=5-\-x,  which  would  give 
y—  6,  7,  8,  9,  10,  or  11. 

\ 

Example  II. 


To  find  a number  such,  that  ^ of  it,  diminished  by  y of  it,  and  by 
5.  shall  be  equal  to  j ^ °f  excess  of  5 times  the  number  above  60 
The  equation  of  this  problem  will  be 

3a;  x 5x—  60 
T~3~5^  12  . 

Clearing  the  equation  of  its  fractions, 

9a; — 4x — 60=5.r — 60  ; 
or  5x — 60=5x  — 60. 

This  last  is  an  identical  Equation,  which  will  be  satisfied  by  attri- 
buting to  x any  numerical  value  whatever.  The  problem  is  therefore  . 
entirely  indeterminate. 


We  may  obtain  an  expression  for  the  value  of  x from  the  last 
equation.  lEus,  by  transposition, 

5x  — 5x—  60  — 60. 

By  adding  similar  terms,  and  retaining  x as  a symbol  in  the  farst 
member,  we  liave 

0.r=0 ; 
which  gives  af=§. 

Hence  § is  a symbol  of  an  indeterminate  quantity. 


The  same  thing  will  appear  from  considering  that  the  quotient  cf 
0 -r~  0 is  any  quantity  ichatever ; inasmuch  as  the  divisor  Ox  any 
quantity  will  produce  the  dividend  0,  (43). 


GENERAL  DESCRIPTION  OF  PROBLEMS. 


207 


3?  Impossible  Problems. 

(272.)  An  Impossible  Problem  is  one  in  which  there  is  some  condi- 
tion, expressed  or  implied,  which  cannot  be  fulfilled. 

An  Impossible  Problem  is  represented  by  a greater  number  of 
independent  equations  than  there  are  unknown  quantities  to  be  deter- 
mined ; or  by  an  equation  in  which  the  value  of  the  unknown  quantity 
is  negative — zero — infinite — or  im, aginary . 

We  subjoin  an  example  of  each  of  these  forms  of  impossibility 

Example  I . 

To  find  two  numbers  whose  sum  shall  be  10,  difference  2,  and  pro- 
duct 20. 

Representing  the  two  numbers  by  x and  y,  we  shall  have 
x-\-y=  10;  x — y — 2;  xy= 20. 

From  the  first  and  second  equations  the  values  of  x and  y will  be 
found  to  be  x = 6,  and  y — 4.  The  third  equation  cannot,  therefore,  be 
fulfilled  ; that  is,  the  problem  is  impossible. 

If  the  third  equation  were  xy=  24,  the  problem  would  be  possible, 
but  this  would  not  be  an  independent  equation,  since  it  may  be  derived 
from  the  other  two. 

Thus,  squaring  the  first  and  second  equations,  and  subtracting,  we 
4au/=96,  or  xy= 24. 


Example  II. 

To  find  a number  which,  added  to  17  and  to  53,  will  make  the 
first  sum  equal  to  ^ of  the  second. 

If  x represent  the  number,  the  equation  will  be 

17_L  53+x 

t7  + a;=___. 


From  this  equation  we  shall  find  x= — 5.  This  number,  added  to 
17  and  53,  gives  12  and  48,  and  12=i  of  48. 

The  problem  is  impossible  in  an  arithmetical  sense,  according  to 
which  addition  always  implies  augmentation ; and  it  is  in  this  sense 
only  that  the  problem  would  be  considered. 

To  make  it  arithmetically  consistent,  it  should  be  stated  thus  : 

To  find  a number  which,  subtracted  from  17  and  from  53,  will  make 
the  first  remainder  equal  to  ~ of  the  second. 


208 


GENERAL  DESCRIPTION  OF  PROBLEMS. 


Example  III.  • 


To  find  a number  such,  that  if  83  he  increased  by  3 times  that 
number,  £ of  the  sum  will  be  equal  to  13f . 


If  x represent  the  number,  the  equation  will  be 
82+3* 

6 ; 


: 1 3j  - 


Clearing  the  equation  of  its  fractions,  we  find 
82  + 3a;=82; 

which  gives  3a?=82  — 82  = 0 ; 

and  a?=^=0,  (50). 

Hence  no  number  can  be  found  that  will  fulfil  the  conditions  of  the 
problem  ; that  is,  the  problem  is  impossible.  The  result  shows  that 
of  82  itself  is  equal  to  13|-. 


Example  IV. 


To  find  a number  such,  that  the  sum  of  \ of  it  and  §-  of  it,  dimi- 
nished by  2,  shall  be  equal  to  of  it  increased  by  3. 

If  x represent  the  number,  the  equation  will  be 
x 2x  „ 11a;  „ 

4+T~2=72+3- 


From  this  equation  we  shall  find 

3x-\-8x  — 24  = llx-\-3&-,  or  11a; — lla;=0a;=60  ; 
and  a?=60°  = ao  , infinity ; (50). 

The  result  shows  that  it  would  require  a number  infinitely  great, 
to  fulfil  the  conditions  of  the  problem.  The  problem  is  therefore 
possible. 


Example  V. 

To  divide  the  number  24  into  two  such  parts,  that  their  product 
shall  be  150. 

If  x represent  one  of  the  two  parts,  24 — x will  represent  the  other 
and  the  equation  will  be 

24x— x2  = 150 
or  x2 — 24z=  — 150,  (117); 
which  gives  ;r=  12± -\/l44 — 150, 

= 12±i/=6. 

In  this  value  of  x,  the  part  V — 6 is  imaginary,  that  is,  it  is  an 
impossible  quantity,  (246) ; hence  the  problem  is  impossible. 


GENERAL  DESCRIPTION  OF  PROBLEMS. 


209 


That  the  preceding  problem  is  impossible,  will  also  appear  from  the 
following  general  proposition ; viz : 

(273-)  The  square  of  one  half  of  any  quantity  is  greater  than  the 
product  of  any  tivo  unequal  parts  of  the  quantity. 

Let  s represent  any  number,  and  d the  difference  between  any  two 
parts  of  the  number ; then  the  respective  parts  are 

o I . fl  e rl  p2 /72 

■ --  and  ^ (168),  and  their  product  is  — - — 

This  product  will  vary  directly  as  its  numerator  s 2 — cl 2 (147);  and 
will  therefore  have  its  greatest  value  when  <7=0.  In  that  case  the  pro- 
duct becomes  j.s2,  which  is  the  square  of  f s , half  the  given  number. 

It  may  also  be  remarked  here,  that  the  sum  of  the  two  equal  factors 
of  any  quantity,  is  less  than  the  sum  of  any  two  unequal  factors  into 
which  the  quantity  can  be  resolved. 

For,  s representing  the  sum,  and  d the  difference,  of  the  two  factors 
of  a quantity,  those  factors  are 


S—rp-  and 4  S ^ - (168),  and  their  product  is 


s2 — d2 
4 


This  product  will  retain  a constant  value , if  s2  and  d 2 be  equally  di- 
minished,i for  then  the  value  of  s2 — d2  will  remain  constant.  In  this 
diminution,  s 2 will  have  its  least  value  when  d2  is  made  0 ; so  that  s, 
the  sum  of  the  two  factors,  will  have  its  least  value  when  a — 0,  and 
the  two  factors  will  then  be  equal  to  each  other. 


Signification  of  the  Different  Forms  under  which  the  Value  of  the 
Unknown  Quantity  may  he  found  in  an  Equation. 


(274.)  1-  Positive  values  of  the  unknown  or  required  quantities,  ful- 
fil the  conditions  of  problems  in  the  sense  in  which  they  are  proposed. 


2.  A value  of  the  unknown  quantity  of  the  form  -j),  shows  that  the 
problem  from  which  the  equation  was  derived  is  indeterminate. 


3.  A negative  value  of  the  unknown  quantity,  in  an  equation  of 
the  first  degree,  indicates  an  impossibility  in  the  problem,  produced  by 
taking  this  quantity  adclitively,  instead  of  subtractively,  or  vice  versa. 

4.  When  the  value  of  the  unknown  quantity  in  an  equation  is  zero. 
\ infinite,  or  imaginary,  the  problem  from  which  the  equation  was  de- 

prived is  impossible. 


10* 


MISCELLANEOUS  PROBLEMS. 


MISCELLANEOUS  PROBLEMS- 

1.  A,  B,  and  C together  have  $2000.  B has  $100  less  than  twice 
as  much  as  A,  and  C $400  less  than  twice  as  much  as  the  other  two 
together,  What  sum  has  each? 

Ans.  A,  $300;  B,  $500  ; C,$1200. 

2.  A gentleman  has  three  plantations.  The  first  contains  250  acres, 
the  second  as  much  as  the  first  and  of  the  third,  and  the  third  as 
much  as  the  first  and  second.  What  is  the  whole  number  of  acres  ? 

Ans.  1200  acres. 

3.  A company  of  workmen  had  been  employed  on  a piece  of  work 

for  24  days,  and  had  half  finished  it,  when,  by  calling  in  the  assistance 
of  16  more  men,  the  remaining  half  was  completed  in  16  days.  What 
was  the  original  number  of  men  ? Ans.  32  meu 

4.  A’s  money  was  equal  to  f of  B’s.  A paid  away  $50  less  than 

| of  his,  and  B $50  more  than  of  his,  when  it  was  found  that  the 
latter  had  remaining  only  A as  much  as  the  former.  What  sum  had 
each  at  first?  Ans.  A,  $300  ; B,  $400. 

5.  A person  wishing  to  enclose  a piece  of  ground  with  palisades,  • 
found,  that  if  he  set  them  one  foot  asunder,  he  would  not  have  enough 
by  150,  but  if  he  set  them  one  yard  asunder,  he  would  have  too  inauv 
by  70.  What  was  the  number  of  his  palisades  ? 

Ans.  180  palisades. 

6.  From  two  tracts  of  land  of  equal  size,  were  sold  quantities 

in  the  proportion  of  3 to  5.  If  150  acres  less  had  been  sold  from  the 
one  which  is  now  the  smaller  of  the  two,  only  |-  as  much  would  have 
been  taken  from  it  as  from  the  other  ; how  many  acres  were  sold  from 
each?  Ans  150,  and  250  acres. 

7.  A and  B had  adjoining  farms,  which,  in  quantity,  were  in  the 

ratio  of  4 to  5.  A sold  to  B 50  acres,  and  afterwards  purchased  from 
B one-third  of  his  entire  tract,  when  it  was  found  that  the  original 
ratio  of  their  quantities  of  land  had  been  reversed.  How  many  acres 
had  each  at  first?  Ans.  A,  200  ; B,  250  acres. 

8.  A waterman  can  row  down  the  middle  of  the  stream,  on 

a certain  river,  5 miles  in  | of  an  hour;  but  it  takes  him  ll  hours  to 
return,  though  he  keeps  along  shore,  where  the  current  is  but  half  as 
strong  as  in  the  middle.  What  is  the  velocity  of  the  middle  of  the 
stream  ? Ans.  2§  miles  per  hour. 

9.  A farmer  has  three  flocks  of  sheep,  whose  numbers  are  in  the 

proportion  of  2,  3,  and  5.  If  he  sell  20  from  each  flock,  the  whole 
number  will  be  diminished  in  the  proportion  of  4 to  3 ; how  many  has 
he  in  each  flock?  H/zs  48,  72,  and  120  sheep 


MISCELLANEOUS  PROBLEMS. 


10.  The  sum  of  $1170  is  to  be  divided  between  three  persons,  A, 
B,  and  0,  in  proportion  to  their  ages.  Now  A’s  age  is  to  B’s  as  1 to 
lg-,  and  to  C’s  as  1 to  2 ; what  are  the  respective  shares  ? 

Ans.  $270  ; $360  ; $540. 

11.  A hare  is  50  leaps  before  a greyhound,  and  takes  4 leaps  to  the 
greyhound’s  3 ; but  2 of  the  greyhound’s  leaps  are  as  much  as  3 of  the 
hare's.  How  many  leaps  must  the  greyhound  take  to  catch  the  hare  ? 

Ans.  300. 

12.  A vintner  has  two  casks  of  wine,  the  contents  of  which  are  in 
the  proportion  of  5 to  6,  and  if  -g-  of  the  quantity  in  the  second  were  to 
be  drawn  off,  the  contents  of  the  two  casks  would  be  equal.  How 
many  gallons  are  there  in  each  ? 

Ans.  This  problem  is  indeterminate;  how  is  its  indeterminate- 
ness indicated  ? 

13.  A person  looking  at  his  watch,  and  being  asked  what  o’clock  it 
was,  replied  that  it  was  between  eight  and  nine,  and  that  the  hour  and 
minute  hands  were  exactly  together.  What  was  the  time  ? 

Ans.  43 m.  38^-s.  past  eight. 

14.  A criminal  having  escaped  from  prison,  traveled  10  hours  be- 
fore his  escape  was  known.  He  was  then  pursued,  and  gained  upon  3 
miles  an  hour.  When  his  pursuers  had  been  S hours  on  the  way  they 
met  an  express  going  at  the  same  rate  as  themselves,  who  had  met  the 
criminal  2 hours  and  24  minutes  before.  In  what  time  from  the  com- 
mencement of  the  pursuit  will  the  criminal  be  overtaken  ? 

Ans.  20  hours. 

15.  A regiment  of  militia  containing  875  men  is  to  be  raised  from 
three  counties,  A,  B,  and  C.  The  quotas  of  A and  B are  in  the  pro- 
portion of  2 and  3,  and  of  B and  C in  the  proportion  of  4 to  5.  What 
is  the  number  to  be  raised  by  each  ? 

Ans.  200,  300,  and  375  men. 

16.  If  19  pounds  of  gold,  in  air,  weighs  18  pounds  in  water;  10 
pounds  of  silver,  in  air,  weighs  9 in  water  ; and  a mass  of  106  pounds, 
composed  of  gold  and  silver,  weighs  99  pounds  in  water ; what  are  the 
respective  quantities  of  gold  and  silver  in  the  mass  ? 

Ans.  76,  and  30  pounds. 

17.  A farmer  having  mixed  a certain  quantity  of  corn  and  oats, 

found  that  if  he  had  taken  6 bushels  more  of  each,  there  would  have 
been  7 bushels  of  corn  to  6 of  oats  ; but  if  he  had  taken  6 bushels  less 
of  each,  there  would  have  been  6 bushels  of  corn  to  5 of  oats.  How 
many  bushels  of  each  were  mixed?  Ans.  78,  and  66  bushels. 

18.  Two  persons,  A and  B,  can  perform  a piece  of  work  in  16  days. 

They  work  together  for  4 days,  when  A being  called  off,  B is  left  to 
finish  it,  which  he  does  in  36  days  more.  In  what  time  could  each  do 
it  separately  ? Ans.  24,  and  48  days. 


K’ 


MISCELLANEOUS  PROBLEMS. 


19.  A merchant  has  two  casks  containing  unequal  quantities  of 
■wine.  Wishing  to  have  the  same  quantity  in  each,  he  pours  from  the 
first  into  the  second  as  much  as  the  second  contained  at  first ; then  he 
pours  from  the  second  into  the  first  as  much  as  was  left  in  the  first ; 
and  then  again  from  the  first  into  the  second  as  much  as  was  left  in  the 
second,  when  there  are  found  to  be  16  gallons  in  each  cask.  How  many 
gallons  did  each  cask  contain  at  first?  Ans.  22,  and  10  gallons. 

20.  A fisherman  being  asked  how  many  fish  he  had  cauuht.  re- 
plied, If  5 be  added  to  one-third  of  the  number  that  I caught  yesterday, 
it  will  make  half  the  number  I have  caught  to-day  ; or  if  5 be  sub- 
tracted from  three  times  this  half,  it  will  leave  the  number  I caught 
yesterday.  How  many  were  caught  each  day? 

Ans.  This  problem  is  impossible ; how  is  its  impossibility 
indicated  ? 

21.  A laborer  engaged  for  n days,  on  condition  that  he  should  re- 
ceive p pence  for  each  day  that  he  worked,  and  forfeit  q pence  for  each 
day  that  he  idled.  At  the  end  of  the  time  he  received  s pence  ; how 
many  days  did  he  work  ? and  how  many  was  he  idle  ? 

Ans.  Worked— ±-  ; was  idle days. 
P+<1  P+<i 


22.  A,  B,  and  C engage  in  a joint  speculation.  A invests  $2000 
for  5 months,  B $2400  for  4 months,  and  C $1600  for  7 months,  The 
profits  amount  to  $4620  ; what  is  each  man’s  share  of  profit  ? 

Ans.  $1500,  $1440,  $1680. 

23.  A farmer  wishes  to  mix  rye  worth  40  cents  a bushel,  and  oats 

worth  26§-  cents  a bushel,  in  such  quantities  as  to  produce  100  bushels 
which  shall  be  worth  30  cents  a bushel.  What  quantity  of  each  must 
be  taken?  Hus.  25,  and  75  bushels. 

24.  Three  persons  engage  in  a joint  mercantile  adventure,  in  which 
the  first  has  the  capital  a for  the  time  b,  the  second  the  capital  c for 
the  time  d,  and  the  third  the  capital  e for  the  time  f.  Their  profits 
amount  to  s ; what  is  each  partner’s  share  of  profit  ? 

abs  cds  efs 

ab-\-cd-{-ef  ’ ab-\-cd-\-ef  ’ ab+ed+ef 

25.  A church  which  cost  $40,000  is  insured,  annually,  at  14  per 
cent.,  for  such  an  amount,  that,  in  case  of  its  being  destroyed  by  tire, 
the  Insurance  Company  shall  be  liable  for  the  cost  of  the  edifice,  and 
the  premium  of  insurance.  What  is  the  sum  insured  ? 

Ans.  40609.13’. 

26.  Four  towns  are  situated  in  the  order  of  the  first  four  letters  ol 


the  alphabet.  The  distance  from  A to  D is  34  miles  ; the  distance 
from  A to  B is  to  the  distance  from  C to  D as  2 to  3 ; and  of  the 
distance  from  A to  B added  to  half  the  distance  from  C to  D.  is  3 times 
the  distance  from  B to  C What  are  the  respective  distances 

Hus  12  4,  and  1$  miles 


MISCELLANEOUS  PROBLEMS. 


27.  A commission  merchant  receives  the  sum  of  s dollars  to  invest 
in  merchandise — himself  to  retain  a commission  of  r per  cent,  on  the 
amount  of  the  purchase.  What  is  the  sum  to  he  invested  ? 


Ans. 


100s 

iOO+r 


28.  A sum  of  money  was  equally  divided  among  a number  of  per- 
sons, by  first  giving  to  A $100  and  g-  of  the  remainder,  then  to  B $200 
and  1 of  the  remainder,  then  to  C $300  and  g of  the  remainder ; and 
so  on.  What  was  the  sum  divided  ? and  the  number  of  persons  ? 

Ans.  $2500,  and  5 persons. 

29.  The  sum  of  $500  is  to  be  applied  in  part  towards  the  payment 

of  a debt  of  $900,  now  due,  and  in  part  to  paying  the  interest,  at  7 per 
cent.,  in  advance,  on  the  remainder  of  the  debt,  on  which  a credit  of 
12  months  is  to  be  allowed.  What  is  the  amount  of  payment  that  can 
be  made  on  the  debt?  Ans.  469.89’ 

30.  A besieged  garrison  had  such  a quantity  of  bread  as  would,  if 
distributed  to  each  man  at  10  ounces  a day,  last  6 weeks  ; but  having  lost 
1200  men  in  a sally,  the  governor  was  enabled  to  increase  the  allow- 
ance to  12  ounces  per  day.  What  was  the  original  number  of  men  ? 

Ans.  7200  men. 

31.  A is  indebted  to  B the  sum  of  s dollars,  and  is  able  to  raise  but 
a dollar’s.  With  this  latter  sum  A proposes  to  pay  a part  of  the  debt, 
and  the  interest,  at  r per  cent.,  in  advance,  on  his  note  at  n years,  for 
the  remainder.  For  what  sum  should  the  note  be  drawn 


Ans. 


100  (s—a) 
100—  rn  ■ 


32.  The  crew  of  a ship  consisted  of  her  complement  of  sailors  and  a 

number  of  soldiers.  Now  there  were  22  sailors  to  every  three  guns, 
and  10  over.  Also  the  whole  number  of  men  was  5 times  the 
number  of  soldiers  and  guns  together.  But  after  an  engagement,  in 
which  the  slain  were  ^ of  the  survivors,  there  wanted  5 of  being  13 
men  to  every  2 guns  Required  the  number  of  guns,  soldiers,  and 
sailors.  Ans.  90  guns,  55  soldiers,  670  sailors. 

33.  Two  sums  of  money,  amounting  together  to  $600,  were  put  at 
interest — the  smaller  at  2 per  cent,  more  than  the  other.  The  interest 
of  the  larger  sum  was  afterwards  increased,  and  that  of  the  smaller  di- 
minished, 1 per  cent.  By  this  the  interest  of  the  whole  was  augmented 
one-twentieth.  But  if  the  interest  of  the  greater  sum  had  been  so  in- 
creased, without  any  diminution  of  the  other,  the  interest  of  the  whole 
would  have  been  increased  one-tenth.  What  were  the  two  sums  ? and 
the  two  rates  of  interest  ? 

Ans.  $400  at  6 per  cent. ; $200  at  8 per  cent. 

34.  Two  persons  purchase  300  acres  of  land  at  $2  per  acre,  each 
one  paying  $300  ; but  the  first  takes  the  more  fertile  portion  of  the 


MISCELLANEOUS  PROBLEMS. 


tract,  at  25  cents  above  the  mean  price  per  acre,  and  the  second  the 
remainder  at  25  cents  below  the  mean  price  per  acre.  How  many 
acres  has  each  ? 

Ans.  This  problem  is  impossible  ; how  is  its  impossibility  indicated  ? 

35.  The  area  of  a field  is  432  square  rods,  and  the  surnof  its  length 

and  breadth  is  equal  to  twice  their  difference.  Required  the  length 
and  breadth  of  the  field.  Ans.  3G,  and  12  rods. 

36.  A and  B lay  out  some  money  on  speculation.  A disposes  ol 
his  interest  in  the  business  for  <£ll,  and  gains  as  much  per  cent,  as 
B lays  out  ; B gains  £36,  and  it  appears  that  A gains  4 times  as  much 
per  cent,  as  B.  What  was  the  capital  of  each  ? Ans.  £5,  and  £120 

37.  A garden  which  is  12  rods  in  length,  and  8 rods  in  breadth,  is 

surrounded  by  a walk  whose  area  is  equal  to  A of  the  area  ol'  the  garden 
itself.  Required  the  breadth  of  the  walk.  Ans  5-f- \/29. 

38.  A and  B hired  a pasture,  into  which  A put  4 horses,  and  B as 

many  as  cost  him  18  shillings  a week.  Afterwards  B put  in  two  addi- 
tional horses,  and  found  that  he  must  pay  20  shillings  a week.  At  what 
rate  was  the  pasture  hired  ? Ans.  30<s.  per  week. 

39.  A gentleman  bought  a rectangular  piece  of  ground,  at  S10  for 

every  rod  in  its  'perimeter . If  the  same  area  had  been  in  the  form  of 
a square,  and  had  been  purchased  in  the  same  way,  it  would  have  cost 
$20  less  ; and  a square  piece  of  the  same  perimeter  would  have  con- 
tained 12-A  square  rods  more.  What  were  the  length  and  breadth  of 
the  lot?  Ans.  16,  and  9 rods. 

40.  A person  being  asked  the  ages  of  himself  and  his  wife,  replied, 

that  the  product  of  their  ages  added  to  the  square  of  his  age,  would 
make  1560.  but  added  to  the  square  of  hers  would  make  12  44.  Yf  hat 
were  their  ages  ? Hras.  30,  and  22. 

41.  A and  B purchased  a farm  containing  900  acres  for  which  they 
paid  $900  each.  On  dividing  the  land,  it  was  agreed  that  A should 
have  his  choice  of  situation,  and  pay  45  cents  per  acre  more  than  B. 
How  many  acres  should  each  have  taken  ? and  at  what  price  per  acre  ? 

Ans.  A 400  acres  at  $2,25  ; B 500  acres  at  $l.s0. 

42.  A capital  of  $13,000  was  divided  into  two  parts,  which  were 
put  at  interest  in  such  a manner  that  the  income  was  the  same  from 
each.  If  the  first  part  had  been  at  the  same  rate  of  interest  as  the 
second,  it  would  have  produced  an  income  of  $360  ; and  if  the  second 
part  had  been  at  the  same  rate  as  the  first,  it  would  have  produced  an 
income  of  $490.  What  were  the  two  rates  of  interest? 

Ans.  7 and  6 per  cent. 

43.  A departs  from  London  towards  Lincoln  at  the  same  time  at 
which  B leaves  Lincoln  for  London.  AVhen  they  met,  A had  traveled 
20  miles  more  than  B,  having  gone  as  far  in  6|-  days  as  B had  in  all 
the  time;  and  it  appeared  that  B would  not  reach  London  under  2 5 


MISCELLANEOUS  PROBLEMS 


days.  What  is  the  distance  between  the  two  places  ? and  how  far 
had  each  man  traveled  ? 

Ans.  Distance,  100  miles ; A had  gone  60,  B 40  miles. 

44.  The  product  of  the  two  dimensions  of  a rectangular  piece  of 
land,  subtracted  from  the  square  of  the  greater  dimension,  leaves  300 
square  rods,  and  subtracted  from  the  square  of  the  less,  leaves  200 
square  rods.  What  are  the  dimensions  of  the  piece  ? 

Ans.  This  problem  is  impossible;  how  is  its  impossibility 
indicated  ? and  in  what  does  this  impossibility  consist  ? 

45.  A person  being  asked  the  ages  of  his  two  children,  replied,  that 
the  difference  of  their  ages  was  3 years,  and  the  product  multiplied  by 
the  sum  of  their  ages  was  308.  What  were  their  ages  ? 

This  problem  will  result  in  an  Affected  Cubic  Equation. 

Ans.  7 and  4 years. 

46.  A gentleman  who  had  a square  lot  of  ground,  reserved  10 

square  rods  out  of  it,  and  sold  the  remainder  for  $432,  which  was  as 
many  dollars  per  square  rod  as  there  were  rods  in  a side  of  the  whole 
square  1 What  was  the  length  of  its  sides  ? Ans.  8 rods. 

47.  A and  B set  out  together  from  the  same  place,  and  travel  in  the 
same  direction.  A goes  the  first  day  28  miles,  the  second  26,  and  so 
on,  in  arithmetical  progression  ; while  B goes  uniformly  20  miles  per 
day.  In  how  many  days  will  the  two  be  together  again  ? 

Ans.  9 days. 

48.  A farmer  wishes  to  build  a crib  whose  capacity  shall  be  1620 
cubic  feet,  and  whose  length,  breadth,  and  height  shall  be  in  an  arith- 
metical progression  decreasing  by  the  common  difference  3.  What 
must  be  the  dimensions  of  the  crib  ? 

It  may  be  well  to  remind  the  student  here,  that  cubic  measure, 
or  measure  of  capacity,  is  found  by  multiplying  together  length , breadth, 
and  height  or  depth.  Ans.  15,  12,  and  9 feet. 

49.  One  traveler  sets  out  to  go  from  A to  B,  at  the  same  time  at  which 

another  sets  out  from  B to  A.  They  both  travel  uniformly,  and  at  such 
rates,  that  the  former,  4 hours  after  their  meeting,  arrives  at  B,  and  the 
latter  at  A,  in  9 hours  after.  In  how  many  hours  did  each  one  per- 
form the  journey?  Ans.  10.  and  15  hours. 

50.  A lady  on  being  asked  the  ages  of  her  three  little  boys,  answered 
that  they  were  in  harmonical  progression  ; and  the  sum  of  their  ages 
was  22  years  ; and  that  if  the  ages  of  the  two  elder  were  each  in- 
creased by  -g-  of  itself,  the  three  would  then  be  in  geometrical  progres- 
sion. What  were  the  respective  ages?  Ans.  4.  6,  and  12  years 

51  A person  wishes  to  construct  two  cubical  reservoirs  which  shall 
differ  in  their  linear  dimensions  by  4 feet,  and  which  shall  together 
contain  5824  cubic  feet.  What  must  be  the  dimensions  of  the  two 
reservoirs?  Ans.  12,  and  16  feet 


MISCELLANEOUS  PROBLEMS. 


52.  Two  partners,  A and  B,  divided  their  gain,  which  was  §60, 

when  B’s  share  was  found,  to  be  $20.  A’s  capital  was  in  trade  4 
months  ; and  if  the  number  50  be  divided  by  A’s  capital,  the  quotient 
will  be  the  number  of  months  that  B’s  capital,  which  was  $100,  con- 
tinued in  trade.  What  was  A’s  capital  ? and  the  time  B’s  was  in 
trade  ? Ans.  A’s  Capital  $50  ; B’s  1 month  in  trade. 

53.  Let  there  be  a square  whose  side  is  110  inches  ; it  is  required 
to  assign  the  length  and  breadth  of  a rectangle  whose  perimeter  shall 
be  greater  than  that  of  the  square  by  4 inches,  but  whose  area  shall 
be  less  than  the  area  of  the  square  by  4 square  inches. 

Ans.  126,  and  96  inches. 

54.  There  is  a number  consisting  of  three  digits  which  increase 

from  left  to  right  by  the  common  difference  2 ; and  the  product  of  the 
three  digits  is  105.  Required  the  number.  Ans.  357. 

55.  A person  bought  2 pieces  of  cloth  for  $63.  For  the  first  piece 
he  paid  as  many  dollars  per  yard  as  there  were  yards  in  both  pieces, 
and  for  the  second  as  many  dollars  per  yard  as  there  were  yards  in 
the  first  more  than  in  the  second  ; also  the  first  piece  cost  six  times  as 
much  as  the  second.  What  was  the  number  of  yards  in  each  piece  ? 

Ans.  6,  and  3 yards. 

56.  There  is  a number  consisting  of  4 digits  which  decrease  from 
left  to  right  by  the  common  difference  2 ; and  the  product  of  the  four 
dibits  is  945.  Required  the  number. 

Ans.  9753. 

57.  A gentleman  purchased  two  square  lots  of  ground  for  $300  ; 
each  of  them  cost  as  many  cents  per  square  rod  as  there  were  rods  in 
a side  of  the  other,  and  the  greater  contained  500  square  rods  more 
than  the  less.  What  was  the  cost  of  each  lot  ? 

Ans.  $180,  and  $120. 

58.  A merchant  bought  a number  of  bales  of  cloth.  The  number 

of  pieces  in  each  bale  was  10  more  than  the  number  of  bales,  and  the 
number  of  yards  in  each  piece  was  5 more  than  the  number  of  pieces 
in  each  bale  ; and  the  whole  quantity  was  1500  yards.  W hat  was 
the  number  of  bales  ? Ans.  5 bales. 

59.  A person  dies,  leaving  children,  and  a fortune  of  S46S0U, 
which,  by  his  will,  is  to  be  divided  equally  amongst  them.  Immedi 
ately  after  the  death  of  the  father,  two  of  the  children  also  die,  in 
consequence  of  which  each  surviving  one  receives  $1950  more  than  he 
was  entitled  to  by  the  will.  How  many  children  did  the  father  leave  ? 

Ans.  8 children. 

60.  A coach  set  out  from  Cambridge  for  London  with  4 more  out- 
side than  inside  passengers.  Seven  outside  passengers  went  at  2 
shillings  less  than  4 inside  ones,  and  the  fare  of  the  whole  amounted 
to  £9.  At  the  end  of  half  the  journey.  3 more  outside  and  one  more 
inside  passenger  were  taken  up,  in  consequence  of  which  the  fare 


MISCELLANEOUS  PROBLEMS. 


the  whole  was  increased  in  the  proportion  of  19  to  15.  Required  the 
number  of  passengers  at  first,  and  the  fare  of  each. 

Ans.  5 inside,  and  9 outside  passengers ; fares  18  and  10  shillings. 

61.  There  is  a fraction  which,  inverted  and  hrcreased  by  i,  will  be 
less  than  2 ; and  if  its  numerator  be  increased  by  2,  the  value  of  the 
fraction  will  be  greater  than  4.  Wliat  is  the  fraction  ? 

Ans.  Between  ? and  J. 

62.  In  a purse  which  contains  24  corns  of  silver  and  copper,  each 

silver  coin  is  worth  as  many  pence  as  there  are  copper  coins,  each 
copper  coin  is  worth  as  many  pence  as  there  are  silver  coins,  and  the 
whole  is  worth  18  shillings.  How  many  were  there  of  each  kind  of 
corns  ? Ans.  6,  and  IS. 

63.  The  square  root  of  a certain  number  plus  4,  is  less  than  9 ; and 
ten  times  the  square  root  of  the  number  minus  2,  is  greater  than  eight 
times  the  square  root  of  the  number  plus  4.  What  is  the  number  ? 

Ans.  Between  9 and  25. 

64.  A and  B traveled  on  the  same  road,  and  at  the  same  rate, 

from  Huntingdon  to  London.  At  the  50th  mile  stone  from  London, 
A overtook  a drove  of  geese,  which  were  proceeding  at  the  rate  of  3 
miles  in  2 hours  ; and  2 hours  afterwards  met  a wagon  which  was 
moving;  at  the  rate  of  9 miles  hi  4 hours.  B overtook  the  same  drove 
of  geese  at  the  45th  mile  stone,  and  met  the  same  wagon  40  minutes 
before  he  came  to  the  3 1st  .mile  stone.  Where  was  B when  A reached 
London  ? Ans.  25  miles  from  London. 


TEXT  BOOKS 

FOR  SCHOOLS  AND  COLLEGES, 

tfit  LUDING  THE  PRIMARY,  ENGLISH,  AND  CLASSICAL  DEPARTMENT,  NATURA1 
SCIENCE,  GEOGRAPHY,  MATHEMATICS,  BOOK-KEEPING,  ETC. 

PUBLISHED  BY 

PRATT,  OAKLEY  AND  COMPANY, 

NO.  21  MURRAY  STREET,  NEW  YORK. 

»*  It  will  be  noticed  that  most  of  these  works  were  written  by  Teachers  of  toe 
.ghesi  eminence. 


Elements  of  Astronomy  ; with  explanatory  Notes  and  ele- 

cant  Illustrations.  By  John  Brocklesby,  A.  M.,  Professor  in  Trinity  College 

$1  25. 

From  the  Connecticut  Common  School  Journal . 

We  take  pleasure  in  calling  the  attention  of  teachers  and  students  to  this  truly  ex 
;ellent  book.  It  is  not  a milk-and-water  compilation,  without  principles  and  with 
>ut  demonstration.  It  contains  the  elements  of  the  science  in  their  proper  integrity 
tnd  proportions.  Its  author  is  a learned  man  and  a practical  instructor,  as  the 
itithor  of  every  school-book  should  be.  The  style  is  a model  for  a text-book,  com 
lining  in  a high  degree  perspicuity,  precision,  and  vivacity.  In  a word,  it  is  the  very 
test  elementary  work  on  Astronomy  with  which  we  are  acquainted. 

This  notice  is  echoed  by  a large  number  of  academies,  who  are  promptly  intro 
lucing  the  book. 

Elements  of  Meteorology;  designed  for  Schools  and  Ac 

ademies.  By  John  Brocklesby,  A.  M.,  Professor  of  Mathematics  and  Natural 

Philosophy  in  Trinity  College.  Hartford  84  cents. 

The  subject  of  Meteorology  is  of  the  deepest  interest  to  all.  Its  phenomena  every 
•vnere  surround  us,  and  ought  to  be  as  familiarly  known  to  the  scholar  as  his  arith- 
i die  or  philosophy.  This  work  treats  of  Win  * in  general,  Hurricanes,  Tornadoes, 
Wa  er-spouts,  Rain,  Fogs.  Clouds,  Dew,  Snow,  Hail,  Thunder-storms,  Rainbows, 
Haloes.*  Meteorites,  Northern  Lights,  &c. 

ii  lias  proved  highly  satisfactory  in  the  school-roorn,  and  is  now  the  established 
text -book  in  a very  large  number  of  our  best  high  schools  and  academies,  where  the 
natural  sciences  are  taught. 

Ii  is  highly  commended  by  Prof.  Olmsted.  Prof.  Silliman,  Dr.  J.  L Comstock 

'r  L*»e.  of  Pa..  Prof.  Love,  of  Mo.,  and  a host  of  eminent  instructors 


2 


Pratt , Oakley  fy  Co's  Publications . 


Views  of  the  Microscopic  World  ; designed  for  Genera* 

Reading,  and  as  a Hand-book  for  Classes  in  Natural  Sciences.  By  Prof  Brockles 

by  $1  12. 

By  the  aid  of  a powerful  microscope,  the  author  has  given  us  highly  ii  structive 
accounts  of  Infusorial  Animalcules,  Fossil  Infusoria,  Minute  Aquatic  Animals, 
Structure  of  Wood  and  Herbs,  Crystallization,  Parts  of  Insects,  &c.,  <fcc. 

To  those  who  are  necessarily  deprived  of  the  aid  of  a microscope,  and  even  to 
those  who  have  it,  this  is  a most  valuable  work.  It  is  clearly  and  pleasantly  written. 
The  sections  on  the  Animalcules,  Infusoria,  and  Crystallization,  are  very  beautifully 
llustrated  with  large  and  expensive  plates.  The  descriptions  of  the  different  kind* 
these  wonderful  little  animals,  many  of  which  multiply  by  billions  in  a few  hours, 
re  really  very  instructive.  There  is  no  better  e;hool  library  book  in  the  world.  It 
■hould  be  read  by  every  man,  woman  and  child. 

Homan  Physiology  ; designed  for  Colleges  and  the  Higher 

Classes  in  Schools,  and  for  General  Reading.  By  Worthington  Hooker,  M.  D 
Prolessor  of  the  Theory  and  Practice  of  Medicine  in  Yale  College.  Illustrated  witr 
nearly  200  engravings.  $1  25. 

This  is  an  original  w>rk,  and  not  a compilation.  It  presents  the  subject  in  a new 
light,  and  at  the  same  time  embraces  all  that  is  valuable  for  its  purpose  that  could  be 
drawn  from  the  most  eminent  sources.  The  highest  encomiums  are  received  from 
all  quarters  ; a lew  are  subjoined. 

From  Caleb  J.  IIallowell,  A exandria  High  School , Ya. 

Hooker’s  Physiology  was  duly  received.  We  propose  to  adopt  it  as  a tex.  booa 
and  shall  order  in  the  course  of  a fortnight. 

From  the  Boston  Medical  and  Surgical  Journal. 

We  can  truly  say  that  we  believe  this  volume  is  of  great  value,  and  we  hope  that 
ihe  rare  merits  of  the  diligent  author  will  be  both  appreciated  and  patronized 

From  B.  F.  Tewksbury,  Lenoxville , Pa. 

I am  ready  to  pronounce  it  unqualifiedly  the  most  admirable  book  or  work  on  the 
human  system  that  has  fallen  under  my  notice,  and  they  have  not  been  few.  If  any 
one  desires  a complete  and  thorough  elucidation  of  the  great  science  discussed,  the* 
can  nowhere  be  better  satisfied  than  in  the  perusal  of  Dr.  Hooker’s  most  excellent 
work. 

An  Introductory  Work  on  Human  Physiology',  by  Prof 

Hooker,  has  just  been  published,  designed  for  all  persons  commencing  the  study 
Dr.  Hooker’s  works  seem  to  have  taken  their  place  decidedly  at  the  head  of  ah 
treatises  on  the  subject  of  Physiology.  They  are  rapidly  going  into  seminaries  and 
normal  schools  in  all  parts  of  the  country,  and  the  best  institutions  express  theii 
“ delight  at  the  result.’'  60  cents. 

A Comparative  English-German  Grammar  ; bused  on  tlie 

affinity  of  the  two  languages.  By  Prof.  Elias  Peissner,  late  ol  the  University  o* 
Munich,  now  of  Union  College,  Schenectady.  §1.00 

From  the  Neir  York  Churchman. 

• * T all  the  German  Grammars  we  have  ever  examined,  this  is  the  most  modest  anc 
u.|ireiending,  and  yet  t contains  a system  and  a principle  which  is  the  life  of  it,  as 
dear,  as  practical,  as  effective  for  lea  ning  grammar  as  any  thing  we  have  ever  seen 
put  forth,  with  so  much  more  pret  nse  cf  originality  and  show  of  philosopht  1: 
vill  be  found,  too,  we  th  ak,  that  the  author  has  not  only  presented  a new  idea  oi 
•nuch  interest  in  itself,  bu  has  admirably  carried  it  out  in  the  practical  lessons  and 
xercises  of  his  work. 

From  Prof  J.  Foster,  of  Schenectady. 

I nave  examined  Prof.  Peis*uicr's  German  Grammar  with  some  attention,  i»a\e 
marked  with  interest  the  rapic  advancement  of  students  here  usinn  it  as  a text  hnok 
and  have  myself  carefully  tested  t in  the  instrtu  tio  y<  ais  <. 

age.  The  result  is  a conviction  fat  it  is  most  admirably  ad:t|  ted  to  secure  eas* 
pleasant,  and  real  progress,  and  t at  from  no  other  work  which  has  come  under  nr. 
notice  can  so  satisfactory  a known-,  of  the  language  be  obtained  in  a &i  /en  time 


Pratt , Oakley  fy  Co’s  Publications. 


3 


Whitlock’s  Geometry  and  Surveying,  is  a work  for  ad- 
vanced students,  possessing  the  highest  claims  upon  the  attention  of  Mathematical 

Teachers.  $1  50. 

*n  comparison  wuh  other  works  of  the  kind,  it  presents  the  following  advantages. 

1.  A better  connected  and  more  progressive  method  of  geometrizing,  calculated  tfl 
enable  the  student  to  go  alone. 

2.  A fuller,  more  varied,  and  available  practice,  by  the  introduction  of  more  than 
four  hundred  exercises,  arithmetical,  demonstrative,  and  algebraical,  so  chosen  as  to 
be  serviceable  rather  than  amusing,  and  so  arranged  as  greatly  to  aid  in  the  acquisi 
tion  of  the  theory 

3.  The  bringing  together  of  such  a body  of  geometrical  knowledge,  theoretical  and 
practical,  as  every  individual  on  entering  into  active  life  demands. 

4.  A system  of  surveying  which  saves  two-thirds  of  the  labor  required  by  the  ordi 
nary  process. 

This  work  is  well  spoken  of  universally,  and  is  already  in  use  in  some  of  the  best 
Institutions  of  this  country.  It  is  recommended  by  Prof.  Pierce,  of  Cambridge,  Prof. 
Smith,  of  Middletown,  Prof.  Dodd,  of  Lexington,  and  many  other  eminent  mathe- 
maticians. 

From  E.  M.  Morse,  Esq. 

I consider  that  I have  obtained  more  mathematical  knowledge  from  Whitlock's 
Geometry  than  from  all  other  text-books  combined.  Unlike  too  many  treatises  of  » 
similar  nature,  it  is  eminently  calculated  to  make  mathematicians 


PROF.  J.  B.  DODD’S  MATHEMATICAL  SERIES 


COMPRISES 

an  Elementary  and  Practical  Arithmetic $0  45 

High  School  Arithmetic 0 84 

Elements  of  Algebra 0 84 

Higher  Algebra 1 50 

Key  to  Algebra 0 84 

Elements  of  Geometry 1 00 


These  books  are  believed  to  be  unrivaled  in  the  following  particulars: 

1.  The  philosophical  accurateness  with  which  their  topics  are  arranged,  so  as  to 
show  the  mutual  dependence  and  relationship  of  their  subjects. 

2.  The  scientific  correctness  and  practical  convenience  of  their  greatly  improved 
nomenclature. 

3.  The  clear  and  concise  manner  in  which  principles  are  stated  and  explanations 
are  given. 

4.  Brevity  and  completeness  of  rules. 

6.  The  distinctness  with  which  the  true  connection  between  Arithmetic  and  its 
cognate  branches  is  developed. 

6 The  excellent  and  thorough  intellectual  discipline  superinduced. 

RECOMMENDATIONS. 

From  R.  T.  P.  Allen,  Svperintendent  of  Kentucky  Military  Institute. 

Upon  a careful  examination  of  a manuscript  Treatise  on  Arithmetic,  by  Proi 
Dodd,  I find  it  greatly  superior  to  all  others  which  have  come  under  my  notice,  in 
s\ stem,  completeness,  and  nomenclature.  The  arrangement  is  natural,  the  system 
complete,  and  the  nomenclature  greatly  improved.  These  improvements  are  not 
slight;  they  are  fundamental — eminently  worthy  the  attention  of  the  mathematical 
teacher,  and  give  a character  of  unity  to  the  work  which  at  once  distinguishes  it  tram 
ii  others  on  this  subject. 

From  C.  M.  Wright,  Associate  Principal  of  Mount  Palatine  Academy. 

I have  examined  Dodd’s  Arithmetic,  and  am  fully  persuaded  that  it  is  supe.no  io 
an)  other  with  which  I am  acquainted.  I could  speak  in  detail  were  it  necessary , 
hilt  all  that  is  required  to  establish  its  reputation  and  introduction , is  to  natu  »» 
uown  by  teachers 


4 


Pratt , Oakley  fy  Co's  Publications . 


From  M.  S Littlefield,  Grand  Rapids , AficA 

I have  Dodd’s  Higher  Arithmetic,  and  unhesitatingly  pronounce  tf  the  best  wor» 
Tor  advanced  classes  I have  ever  seen. 

From  E.  Hinds,  Esq.,  of  Newtown  Academy. 

i have  recently  adopted  Dodd’s  High  School  Arithmetic,  and  like  it  much.  Having 
seen  that  Prof.  Dodd  is  also  author  of  an  Algebra,  I should  like  to  see  that  work  be 
ore  forming  a new  class. 

From  H.  Elias,  Esq.,  Palmyra , Mo. 

1 have  fairly  tested  Dodd’s  Algebra,  and  am  much  pleased  with  it.  If  I like  liis 

eomeiry  as  well  as  the  Algebra,  I shall  forthwith  introduce  it  into  my  school. 

From  Prof.  W.  H.  De  Puy. 

We  have  introduced  Dodd’s  Algebra  into  the  Genesee  Wesleyan  Seminary  as  s 
permanent  text-book. 

From  R.  H.  Moore,  III. 

Dodd’s  Algebra  possesses  excellencies  pertaining  to  no  other  work 

From  Rev  J.  A.  McCanley,  Va. 

I am  much  pleased  with  Dodd’s  Algebra,  and  will  introduce  it. 

From  Oscar  Harris,  N.  J 

i use  Professor  Dodd’s  Algebra,  and  shall  continue  it  as  our  regular  text-boot 
From  Prof.  A.  L.  Hamilton,  President  of  Andrew  College. 

1 have  examined  with  some  care  Prof.  Dodd’s  Elements  of  Geometry,  and.  so  raj 
ns  1 am  capable  of  judging,  1 conceive  it  to  be  in  many  respects  decidedly  the  best 
work  of  the  kind  extant  For  simplicity,  exactness,  and  completeness,  it  can  have 
no  superior.  Like  his  Arithmetic  and  Algebra,  in  many  important  particulars,  hia 
Geometry  stands  pre-eminent  and  alone. 


A New  Common-School  Arithmetic,  by  Prof.  Dodd,  is  in 

press. 

The  Department  of  Public  Instruction  in  Canada  has  repeatedly  ordered  Prof 
Dodd's  books,  as  well  as  many  of  F.  B.  <fc  Co.’s  other  publications,  for  use  in  schools 

Schell’s  Introductory  Lessons  in  Arithmetic;  designed 

as  an  Introduction  to  the  study  of  any  Mental  or  Wnuen  Arithmetic.  It  contains 
a large  amount  of  mental  questions  together  with  a large  number  of  questions  to 
be  performed  on  the  slate,  thus  combining  mental  and  written  exercises  lor  young 
beginners.  Tins  is  a very  attractive-  little  book,  superior  to  any  of  its  class.  It 
leads  the  pupil  on  by  the  easiest  steps  possible,  and  yet  insures  constant  pro- 
gress. 20  cents. 

From  Geo.  Payne  Quackenbos,  Rector  of  Henry  street  Grammar  School,  N.  V 
It  is  unnecessary  to  do  more  than  to  ask  the  attention  of  teachers  to  this  woik, 
they  cannot  examine  it  impartially  without  being  convinced  of  its  superior  merits 
It  will,  no  doubt,  become  one  of  the  most  popular  of  school-books. 

From  J.  Markham,  Ohio. 

1 wish  to  introduce  Schell’s  little  Arithmetic.  It  is  just  the  thing  for  beginners 
Vend  six  dozen 

From  G.  C.  Merrifield,  hid. 

I ain  highly  pleased  with  Schell’s  little  book,  and  shall  use  it- 

From  D.  F.  Dewolf  Ohio. 

Schell’s  little  book  for  children  is  a beau-ideal  of  my  own,  and  of  course  it  suit* 

From  D.  G.  IIeffron,  Sup't.  Schools , Utica. 

The  School  Committee  have  adopted  Schell's  Arithmetic  for  our  publie 
send  us  three  honored. 


Pratt , Oakley  Co's  Publications. 


* 


An  Intellectual  and  Practical  Arithmetic  ; or,  First 

Lessons  in  Arithmetical  Analysis.  By  L L.  Enos,  Graduate  of  tlie  New  York 
State  Normal  Schools.  25  cents. 

The  same  clearness  and  cojiciseness  characterize  this  admirable  book  that  belong 
to  the  works  of  Prof.  Dodd.  The  natural  arrangements  of  the  text,  and  the  logical 
nude  of  solving  the  questions,  is  a peculiar  and  important  feature  belonging  to  this 
I onok  alone. 

From  Prof.  C.  M.  Wright. 

I have  examined  with  care  and  interest  Enos’  Mental  Arithmetic,  and  shall  intro 
I Alice  it  at  once  into  the  Academy. 

From  Profs.  D.  I.  Pinckney,  S.  M.  Fellows,  S.  Searle,  Rock  River  Seminary 
have  examined  an  intellectual  Arithmetic,  by  J.  L.  Enos,  and  like  it  much 
We  shall  immediately  use  it  in  our  school. 

Prof.  Palmer’s  Book-Keeping  ; Key  and  Blanks.  67  cents 

This  excellent  book  is  superior  to  the  books  generally  used,  because : 

1.  It  contains  a large  number  of  business  blanks  to  be  filled  by  the  learner,  such  as 
deeds,  mortgages,  agreements,  assignments,  &c.,  &c. 

2.  Explanations  from  page  to  page,  from  article  to  article,  and  to  settle  principles 
oflaw  in  relation  to  deeds,  mortgages,  <fcc.,  &c. 

3.  The  exercises  are  to  be  written  out,  after  being  calculated.  In  other  works,  the 
J pupil  is  expected  to  copy,  merely. 

Palmer’s  Book-Keeping  is  used  in  the  New  York  Public  Schools,  and  extensively 
In  Academies,  It  is  n commended  by  Horace  Webster,  LL.  D.,  G.  B.  Docharty, 
LL.  D.,  and  a large  number  of  accountants  and  teachers. 


REV.  P.  BULLIONS’  ENGLISH  AND  CLASSICAL  SERIES, 


COMPRISING 

Practical  Lessons  in  English  Grammar  and  Composition $0  25 

Principles  of  English  Grammar 0 50 

Progressive  Exercises  in  Analysis  and  Parsing 0 15 

Introduction  to  Analytical  Grammar . . 0 30 

New,  or  Analytical  and  Practical  English  Grammar 0 63 

Latin  Lessons,  with  Exercises  in  Parsing.  By  Geo.  Spencer,  A.  M.  Half 

eloth,  enlarged 0 63 

Bullions’ Principles  of  Latin  Grammar ...  100 

Bullions’  Latin  Reader.  With  an  Introduction  on  the  Idioms  of  the  Latin 

Language.  An  improved  Vocabulary  1 00 

Bullions’  OvEsar’s  Commentaries 1 00 

Bullions’  Cicero’s  Orations.  With  reference  both  to  Bullions’,  and  An- 
drew’s, and  Stoddard’s  Latin  Grammar 1 IS 

Bullions’  Sallust 1 00 

Bullions-  Greek  Lessons  for  Beginners 0 75 

Bullions’  Principles  of  Greek  Grammar  1 13 

Bullions’  Greek  Reader  Wit*i  Introduction  on  the  Idioms  of  the  Greek 

Language,  and  Improved  Lexicon 1 75 

Bullions’ Latin  Exercises 1 25 

Cooper’s  Virgil 2 00 


In  tbis  series  of  books,  the  three  Grammars,  English,  Latin,  and  Greek,  are  all  on 
the  same  plan  The  general  arrangement,  definitions,  rules,  &c.,  are  the  same,  and 
expressed  in  the  same  language,  as  nearly  as  the  nature  of  the  case  would  admit 
To  those  who  study  Latin  and  Greek,  much  time  and  labor,  it  is  believed,  will  be 
ewved  I y this  method,  both  to  teacher  and  pupil.  The  analogy  and  peculiarities  of 
• he  different  languages  being  kept  in  view,  will  show  what  is  common  to  ail.  rr  pecu 


8 


Pratt , Oakley  Co's  Publications . 


&ar  to  each  ; the  confusion  and  difficulty  unnecessarily  occasioned  by  the  use  of  ele- 
men t ary  works  differing  widely  from  each  other  in  language  and  structure,  will  b% 
avoided,  and  the  progress  of  the  student  rendered  much  more  rapid,  easy,  and  satis 
factory. 

No  senes  of  Grammars,  having  this  object  in  view,  has  heretofore  been  prepared, 
and  the  advantages  which'  they  offer  cannot  be  obtained  in  an  equal  degree  by  thl 
study  of  any  other  Grammars  now  in  use.  They  form  a complete  course  of  element- 
ary books,  in  which  the  substance  of  the  latest  and  best  Grammars  in  each  language 
nas  been  compressed  into  a volume  of  convenient  size,  beautifully  printed  on  supe 
rior  paper,  neatly  and  strongly  bound,  and  are  put  at  the  lowest  prices  at  which  they 
can  be  afforded. 

The  elementary  works  intended  to  follow  the  Grammars — namely,  the  Latin 
Reader  and  the  Greek  Reader — are  also  on  the  same  plan  ; are  prepared  with  specie! 
references  to  these  works,  and  contain  a course  of  elementary  instruction  so  uniqu; 
and  simple  as  to  furnish  great  facilities  to  the  student  in  these  languages. 

NOTICES. 

From  Prof.  C.  S.  Pennel,  Antioch  College , Ohio. 

Bullions’  books,  by  their  superior  arrangement  and  accuracy,  their  completeness 
as  a series,  and  the  references  from  one  to  the  other,  supply  a want  more  perfectly 
than  any  other  books  have  done.  They  bear  the  marks  of  the  instructor  as  well  as 
the  scholar.  It  requires  more  than  learning  to  make  a good  school-book. 

bVorn  J.  B.  Thompson,  A.  M.,  late  Hector  of  the  Somerville  Classical  Institute , N.  J. 

I use  Bullions’  works — all  of  them — and  consider  them  the  best  of  the  kind  that 
have  been  issued  in  this  or  any  other  language.  If  they  were  universally  used  we 
would  not  have  so  many  superficial  scholars,  and  the  study  ol  the  classics  would  be 
more  likely  to  serve  the  end  for  which  it  was  designed — the  strengthening  and 
adorning  of  the  mind. 

From  A.  C.  Richards,  Esq.,  Clay  Co.,  Ga. 

Wc  think  Bullions’  Latin  Grammar,  in  the  arrangement  of  its  syntax  and  the  ccn- 
•iseuess  of  its  rules,  the  manner  of  treating  prosody,  and  the  conjugal  ions  of  .he 
verbs,  superior  to  any  other.  If  his  Greek  Reader  is  as  good  as  the  Latin  Reader,  we 
shall  introduce  it. 

It  is  almost  superfluous  to  publish  notices  of  books  so  extensively  used. 

Within  the  last  few  months  Dr.  Bullions’  English  Grammar  has  been  introduced 
into  the  Public,  and  many  of  the  Private  Schools,  the  Latin  School,  the  English 
High  School,  the  City  Normal  School,  of  the  city  of  Bost'  ' ; Normal  Schools  of 
Bridgewater  and  Westfield  , Marlborough  Academy  ; cities  Salem.  Newbnryport. 
&c.,  Mass.  ; Portsmouth,  Concord,  and  several  academies!  New  Hampshire;  and 
re-adopted  in  Albany  and  Troy.  New  York.  They  are  used  in  over  seventy  acade 
mies  in  New  York,  ami  in  many  of  the  most  flourishing  institutions  in  every  Slate  of 
the  Union.  Also,  in  the  Public  Schools  of  Washington,  D.  C.,  and  of  Canada.  :n 
Oregon  and  Australia.  The  classical  Series  has  been  introduced  into  several  col 
leges,  and  it  is  not  too  much  to  say  that  Bullions’  Grammars  bid  fair  to  become  the 
Standard  Grammars  of  the  country. 


THE  STUDENTS’  SEEIES 

BY  J.  S.  DENMAN,  A.  .\L 

Cents. 


1 he  Students’  Primer ? 

14  44  Spelling-Book II 

*•  44  First  Reader 13 

44  4 Second  44  2! 

4 44  Third  44  

44  44  Fourth  44  71 

* “ Fifth  44  94 

“ 14  Speaker  V 


Pratt,  Oakley  Co's  Publications . 


7 


Tho  Publishers  feel  justified  in  claiming  that  the  Students’  Series  is  decidedly  the 
vest  for  teaching  readme,  and  spelling  that  has  yet  appeared.  The  plan  of  teaching 
includes,  in  the  first  steps,  an  ingenious  and  original  mode  of  repetition  which  is 
very  pleasing  and  encouraging  to  the  pupil.  The  first  books  of  the  series  are  very 
instructive,  and  the  later  portions  consist  of  fine  selections,  which  are  not  hack- 
neyed. Prof.  Page,  late  Principal  of  the  New  York  State  Normal  School,  said  of  this 
system:  “7?  is  the  best  I ever  saw  for  teaching  the  first  principles  of  Reading.” 
Such  testimony  is  of  the  highest  value,  and  none  need  be  afraid  to  use  the  books  on 
such  a recommendation. 

The  numerous  notices  from  all  parts  of  the  country  where  these  books  have  been 
used,  cannot  be  introduced  here.  They  have  just  gone  into  the  schools  of  Seneca 
County,  N.  Y.,  without  solicitation  ; and  the  same  is  true  of  many  important 
schools  where  they  have  been  examined. 

From  C.  B.  Crumb,  N.  Y. 

The  Students’  Series  is,  in  my  opinion,  the  best  in  use.  I believe  a class  of  young 
students  will  learn  twice  as  much,  with  the  same  labor,  as  they  would  from  any  other 
system.  The  books  of  this  Series  excel  in  the  purity  and  attraction  of  their  style 
1 have  introduced  them. 


DR.  COMSTOCK’S  SERIES  OE  BOOKS  ON  THE  SCIENCES,  viz: 

Introduction  to  Natural  Philosophy.  For  Children $0  43 

System  of  N vtural  Philosophy,  newly  revised  and  enlarged,  including  late 

discoveries 1 (Ml 

Elements  of  Chemistry.  Adapted  to  the  present  state  of  the  Science 1 00 

The  Young  Botanist.  New  edition 50 

Elements  of  Botany.  Including  Vegetable  Physiology,  and  a Description  of 

Common  Plants.  With  Cuts 1 25 

Outlines  of  Physiology,  both  Comparative  and  Human.  To  which  is  added 
Outlines  of  Anatomy,  excellent  for  the  general  scholar  and  ladies’  schools.  80 

New  Elements  of  Geology.  Highly  Illustrated 1 25 

Elements  of  Mineralogy.  Illustrated  with  numerous  Cuts 75 

Natural  History  of  Birds.  Showing  their  Comparative  Size.  A new  and 

valuable  feature 50 

vkvuMM  Htstop-  of  Beasts.  Ditto 50 

Natural  H sto^y  vf  and  Beasts.  Do.  Cloth 1 00 

Questions  and  Illustrations  to  the  Philosophy 30 

All  the  above  works  are  fully  illustrated  by  elegant  cuts. 


The  Philosophy  has  been  republished  in  Scotland,  and  translated  for  the  use  o 
schools  in  Prussia.  The  many  valuable  additions  to  the  work  by  its  transatlantic 
editors,  Prof.  Lees,  of  Edinburgh,  and  Prof,  lloblyn,  of  Oxford,  have  been  embraced 
Dy  the  author  in  his  last  revision.  The  Chemistry  has  been  entirely  revised,  and 
contains  all  the  late  discoveries,  together  with  methods  of  analyzing  minerals  and 
netals.  Portions  of  the  series  are  in  course  of  publication  in  London.  Such  testi- 
mony, in  addition  to  the  general  good  testimony  of  teachers  In  this  country,  is  suffi 
cient  to  warrant  us  in  saying  that  no  works  on  similar  subjects  can  equal  them,  or 
have  ever  been  so  extensively  used.  Continual  applications  arc  made  to  the  publish- 
ers to  replace  the  Philosophy  in  schools  where,  for  a time,  it  has  given  way  to  other 
booke.  The  style  of  Dr.  Comstock  is  so  clear,  and  his  arrangement  is  so  excellent, 
that  no  writer  can  be  found  to  excel  him  for  school  purposes,  and  he  takes  constant 
pains  to  include  new  discoveries,  and  to  consult  eminently  scientific  men. 


HON.  J.  OLNEY’S  GEOGRAPHICAL  SERIES. 

Ppimary  Geography;  with  Colored  Maps.  25  cents. 
Quarto  Geography  ; with  elegant  Cuts,  Physical  Geogra- 

phy  Tables,  Map  of  the  Atlantic  Oceno  &c.  75  cents 


e 


Pratt,  Oakley  fy  Co’s  Publications. 


Olney’s  School  Geography  a\td  Atlas.  Containing  An 

cient  Geography,  Physical  Geography,  Tables,  an  entirely  new  Chari  of  thi 
World,  to  show  its  physical  contonnation,  as  adapted  to  purposes  of  commerce, 
and  also  for  the  purpose  of  reviewing  classes  ; also  a Chronological  Table  of  Disco 
veries.  $1  12. 

)lney’s  Outline  Maps.  Of  the  World,  United  States 

Europe,  Asia,  Africa,  America,  and  Canada,  with  Portfolio  and  Book  of  Exercises 

$6. 

All  the  recent  improvements  are  included  in  Olney’s  Quarto  and  School  Geogra 
hies.  They  are  not  obsolete  or  out  of  date,  but  fully  “ up  to  the  times.”  In  ele« 
ance  or  completeness  they  are  not  surpassed. 

Mr.  Olney  commenced  the  plan  of  simplifying  the  first  lesson,  and  teaching  a child 
by  what  is  familiar,  to  the  exclusion  of  astronomy.  He  commenced  the  plan  of  hav- 
ing only  those  tilings  represented  on  the  maps  which  the  pupil  was  required  to 
learn.  He  originated  the  system  of  classification,  and  of  showing  the  government, 
religion,  &c.,  by  symbols.  He  first  adopted  t lie  system  of  carrying  the  pupil  over 
the  earth  by  means  of  the  Atlas.  His  works  fir-t  contained  cuts,  in  which  the  dress 
architecture,  animals,  internal  improvements,  <fcc.,  of  each  country  are  grouped,  so 
as  to  be  seen  at  one  view.  His  works  Hr«t  contained  the  world  as  known  to  the  An- 
cients. as  an  aid  to  Ancient  History,  a'  i a Synopsis  of  Physical  Geography,  with 
maps.  In  short,  we  have  seen  no  valuable  feature  in  any  geography  which  has  not 
originally  appeared  in  these  works;  and  we  think  it  not  too  much  to  claim  that,  in 
many  respects,  most  other  works  are  copies  of  these.  We  think  that  a fair  and 
candid  examination  will  show  that  Olney’s  Atlas  is  the  largest,  most  systematic, 
and  complete  of  any  yet  published,  and  that  the  Quarto  and  Modern  School  Geogra- 
phies contain  more  matter,  and  that  better  arranged,  than  any  similar  works  ; and 
they  are  desired  to  test  the  claims  here  asserted. 

It  is  impossible  to  give  here  more  than  a fractional  part  of  the  recommendations, 
of  the  first  order,  which  the  publishers  have  received  for  the  foregoing  list  of  books 
Enough  has  been  given  to  show  the  claims  of  the  books  to  examination  and  use. 

All  these  works  are  made  in  very  neat,  durable  style,  ami  are  sold  as  tow  as  a 
moderate  remuneration  will  allow.  Copies  supplied  to  teachers  for  their  own  u>p  at 
ene-fifth  off  from  the  retail  price,  and  postage  paid.  Large  institutions  are  furnished 
•ample  copies  without  charge. 


PRATT,  OAKLEY  & CO. 

21  Murray  Street,  Xew  York. 


